Long training field sequence construction

ABSTRACT

In wireless communications for a 20 megahertz (MHz) channel bandwidth, a first device may determine a high efficiency long training field (HE-LTF) mode. The first device may generate an HE-LTF symbol by using a portion or an entirety of an HE-LTF sequence corresponding to the channel bandwidth and HE-LTF mode. The first device may transmit, in the channel bandwidth, a high efficiency physical layer protocol data unit (HE PPDU) that includes the HE-LTF symbol. A second device may receive, in the 20 MHz channel bandwidth, a downlink HE PPDU that includes an HE-LTF symbol. The second device may obtain, from the HE-LTF symbol, a portion or an entirety of an HE-LTF sequence corresponding to the channel bandwidth and an HE-LTF mode of the HE-LTF symbol. The downlink HE PPDU may be the HE PPDU from the first device. Other methods, apparatus, and computer-readable media are also disclosed.

CROSS-REFERENCE TO RELATED APPLICATIONS

This application is a continuation of application Ser. No. 16/405,933,filed on May 7, 2019, now U.S. Pat. No. 10,945,158, which is acontinuation of application Ser. No. 15/452,567, filed on Mar. 7, 2017,now U.S. Pat. No. 10,327,172, which is a continuation of applicationSer. No. 15/079,007, filed on Mar. 23, 2016, now U.S. Pat. No.9,628,310, which claims the benefit of U.S. Provisional Application No.62/138,302, filed on Mar. 25, 2015, U.S. Provisional Application No.62/157,849, filed on May 6, 2015, U.S. Provisional Application No.62/214,139, filed on Sep. 3, 2015, U.S. Provisional Application No.62/214,156, filed on Sep. 3, 2015, U.S. Provisional Application No.62/236,815, filed on Oct. 2, 2015, U.S. Provisional Application No.62/250,944, filed on Nov. 4, 2015, and U.S. Provisional Application No.62/264,812, filed on Dec. 8, 2015, the entirety of each of which isincorporated herein by reference for all purposes.

TECHNICAL FIELD

The present description relates in general to wireless communicationsystems, and more particularly to, for example, without limitation, longtraining field sequence construction.

BACKGROUND

Wireless local area network (WLAN) devices are deployed in diverseenvironments. These environments are generally characterized by theexistence of access points and non-access point stations. Increasedinterference from neighboring devices gives rise to performancedegradation. Additionally, WLAN devices are increasingly required tosupport a variety of applications such as video, cloud access, andoffloading. In particular, video traffic is expected to be the dominanttype of traffic in many high efficiency WLAN deployments. With thereal-time requirements of some of these applications, WLAN users demandimproved performance in delivering their applications, includingimproved power consumption for battery-operated devices.

The description provided in the background section should not be assumedto be prior art merely because it is mentioned in or associated with thebackground section. The background section may include information thatdescribes one or more aspects of the subject technology.

BRIEF DESCRIPTION OF THE DRAWINGS

FIG. 1 illustrates a schematic diagram of an example of a wirelesscommunication network.

FIG. 2 illustrates a schematic diagram of an example of a wirelesscommunication device.

FIG. 3A illustrates a schematic block diagram of an example of atransmitting signal processor in a wireless communication device.

FIG. 3B illustrates a schematic block diagram of an example of areceiving signal processor in a wireless communication device.

FIG. 4 illustrates an example of a timing diagram of interframe space(IFS) relationships between wireless communication devices.

FIG. 5 illustrates an example of a timing diagram of a carrier sensemultiple access/collision avoidance (CSMA/CA) based frame transmissionprocedure.

FIG. 6 illustrates a very high throughput (VHT) frame format.

FIGS. 7A through 7D illustrate examples of applying a phase rotation toa VHT long training field (LTF) sequence.

FIG. 8 illustrates an example of a high efficiency (HE) frame.

FIG. 9 illustrates example numerology for a 20 MHz channel bandwidth.

FIG. 10 illustrates example numerology for a 20 MHz channel bandwidth.

FIG. 11 illustrates example numerology for a 40 MHz channel bandwidth.

FIG. 12 illustrates example numerology for an 80 MHz channel bandwidth.

FIGS. 13A through 13C and 14A through 14C illustrate examples ofdifferent options for presenting a start tone index and an end toneindex for each of the phase rotation blocks.

FIG. 15 illustrates an example of applying phase rotationhierarchically.

FIGS. 16A through 16C illustrate examples of applying phase rotationshierarchically.

FIGS. 17A through 17D illustrate examples of duplication of a 26-tonesequence for the channel bandwidth 20 MHz resource allocationillustrated in FIG. 10 .

FIGS. 18A through 18D illustrate examples of duplication of a 26-tonesequence for the channel bandwidth 40 MHz resource allocationillustrated in FIG. 11 .

FIG. 19A illustrate an example of an LTF for an 80 MHz channel bandwidthconstructed using LTF sequences for a 20 MHz channel bandwidth.

FIG. 19B illustrate an example of an LTF for an 80 MHz channel bandwidthconstructed using LTF sequences for a 40 MHz channel bandwidth.

FIGS. 20 through 31 illustrate examples of peak-to-average power ratio(PAPR) for various resource allocations.

FIGS. 32 through 34 illustrate examples of center tone optimization.

FIGS. 35A and 35B illustrate examples of an odd subsampling of a26-length sequence.

FIGS. 36A and 36B illustrate examples of an even subsampling of a26-length sequence.

FIGS. 37 through 40 illustrate examples of LTF tone subsampling fordifferent channel bandwidths.

FIG. 41 illustrates an example of a chart for 2×LTF tones associatedwith subsampling.

FIGS. 42 through 44 illustrate examples of LTF transmission for uplinkand downlink.

FIG. 45 shows examples of structures of high efficiency (HE) LTFsequences in 1× HE LTF mode for 20 MHz, 40 MHz, and 80 MHz channelbandwidths.

FIGS. 46 through 52 illustrate flow charts of examples of operationsfacilitating wireless communication.

In one or more implementations, not all of the depicted components ineach figure may be required, and one or more implementations may includeadditional components not shown in a figure. Variations in thearrangement and type of the components may be made without departingfrom the scope of the subject disclosure. Additional components,different components, or fewer components may be utilized within thescope of the subject disclosure.

DETAILED DESCRIPTION

The detailed description set forth below is intended as a description ofvarious implementations and is not intended to represent the onlyimplementations in which the subject technology may be practiced. Asthose skilled in the art would realize, the described implementationsmay be modified in various different ways, all without departing fromthe scope of the present disclosure. Accordingly, the drawings anddescription are to be regarded as illustrative in nature and notrestrictive. Like reference numerals designate like elements.

One or more aspects of the present disclosure relate to increasingefficiency in wireless communications and improving user experience inwireless environments. One or more aspects of the present disclosureprovide techniques for constructing a long training field (LTF) sequencefor a preamble of a high efficiency (HE)-based transmission. Such an LTFsequence may be referred to as an HE LTF sequence. Peak-to-average powerratio (PAPR) may be utilized to evaluate HE LTF sequences, with HE LTFsequences of lower PAPR values generally being more desirable. One ormore aspects of the present disclosure utilizes orthogonalfrequency-division multiple access (OFDMA).

In one or more implementations, HE LTF sequences are constructed forvarious channel bandwidths (e.g., 20 MHz, 40 MHz, 80 MHz, 160 MHz, and80+80 MHz). Phase rotation may be applied to tones (e.g., blocks oftones) within an HE LTF sequence, where one or more reserved tones maybe included into such blocks. Further, hierarchical phase rotation maybe utilized. In one aspect, usable center tones are optimized so thattheir corresponding PAPR values may be lowered. In one aspect, an HE LTFsequence may be subsampled so that a subset of the HE LTF sequence(e.g., about a half of the elements of the HE LTF sequence) istransmitted. In downlink-OFDMA (DL-OFDMA), in some aspects, a station(e.g., an access point) transmits an HE LTF sequence only for thosetones that have been allocated to the stations that participate in theOFDMA communication. In one aspect, an access point transmits an HE LTFsequence for the data/pilot tones allocated to the stations as well asreserved tones. In uplink-OFDMA (UL-OFDMA), each station thatparticipates in the OFDMA communication may transmit a subset of an HELTF sequence, where the subset is associated with the tones that havebeen allocated to the station. In one or more aspects, wirelesscommunications utilizing HE LTF sequences can allow for lower PAPRand/or higher transmission efficiency.

FIG. 1 illustrates a schematic diagram of an example of a wirelesscommunication network. In the wireless communication network 100, suchas a wireless local area network (WLAN), a basic service set (BSS)includes a plurality of wireless communication devices (e.g., WLANdevices). In one aspect, a BSS refers to a set of stations (STAs) thatcan communicate in synchronization, rather than a concept indicating aparticular area. In this example, the wireless communication network 100includes wireless communication devices 111-115, which may be referredto as stations (STAs).

Each of the wireless communication devices 111-115 may include a mediumaccess control (MAC) layer and a physical (PHY) layer according to anInstitute of Electrical and Electronics Engineers (IEEE) 802.11standard. In this example, at least one wireless communication device(e.g., device 111) is an access point (AP). An AP may be referred to asan AP STA or an AP device. The other wireless communication devices(e.g., devices 112-115) may be non-AP STAs. Alternatively, all of thewireless communication devices 111-115 may be non-AP STAs in Ad-hocnetworking.

An AP STA and a non-AP STAs may be collectively called STAs. However,for simplicity of description, in some aspects, only a non-AP STA may bereferred to as a STA. An AP may be, for example, a centralizedcontroller, a base station (BS), a node-B, a base transceiver system(BTS), a site controller, a network adapter, a network interface card(NIC), a router, or the like. An non-AP STA (e.g., a client deviceoperable by a user) may be, for example, a device with wirelesscommunication capability, a terminal, a wireless transmit/receive unit(WTRU), a user equipment (UE), a mobile station (MS), a mobile terminal,a mobile subscriber unit, a laptop, a non-mobile computing device (e.g.,a desktop computer with wireless communication capability) or the like.In one or more aspects, a non-AP STA may act as an AP (e.g., a wirelesshotspot).

In one aspect, an AP is a functional entity for providing access to adistribution system, by way of a wireless medium, for an associated STA.For example, an AP may provide access to the internet for one or moreSTAs that are wirelessly and communicatively connected to the AP. InFIG. 1 , wireless communications between non-AP STAs are made by way ofan AP. However, when a direct link is established between non-AP STAs,the STAs can communicate directly with each other (without using an AP).

One or more implementations of the present disclosure relate to a HighEfficiency (HE) WLAN system in which an HE STA supporting the IEEE802.11ax standard may coexist with one or more non-HE STAs, such as avery high throughput (VHT) STA supporting the IEEE 802.11ac standard, anHT STA supporting the IEEE 802.11n, and/or a legacy STA (e.g., a non-HTSTA supporting the IEEE 802.11a/b/g standard). The IEEE 802.11a/b/g/n/ac standards are incorporated herein by reference. Further, oneor more aspects of the present disclosure relate to OFDMA, which is atechnique that can be used in next generation WLAN technologies such asthe HE technologies (e.g., technologies supporting the IEEE 802.11axstandard). In one or more aspects, the present disclosure relates toWLAN indoor and outdoor operation in the 2.4 GHz and the 5 GHz frequencybands. Additional bands (e.g., bands between 1 GHz and 6 GHz) may beadded as they become available.

FIG. 2 illustrates a schematic diagram of an example of a wirelesscommunication device. The wireless communication device 200 includes abaseband processor 210, a radio frequency (RF) transceiver 220, anantenna unit 230, a memory 240, an input interface unit 250, an outputinterface unit 260, and a bus 270, or subsets and variations thereof.The wireless communication device 200 can be, or can be a part of, anyof the wireless communication devices 111-115.

In this example, the baseband processor 210 performs baseband signalprocessing, and includes a medium access control (MAC) processor 211 anda physical layer (PHY) processor 215. The memory 240 may store software(such as MAC software) including at least some functions of the MAClayer. The memory may further store an operating system andapplications.

In this illustration, the MAC processor 211 includes a MAC softwareprocessing unit 212 and a MAC hardware processing unit 213. The MACsoftware processing unit 212 executes the MAC software to implement somefunctions of the MAC layer, and the MAC hardware processing unit 213 mayimplement remaining functions of the MAC layer as hardware (MAChardware). However, the MAC processor 211 may vary in functionalitydepending on implementation. The PHY processor 215 includes atransmitting (TX) signal processing unit 280 and a receiving (RX) signalprocessing unit 290. The term TX may refer to transmitting, transmit,transmitted, transmitter or the like. The term RX may refer toreceiving, receive, received, receiver or the like.

The PHY processor 215 interfaces to the MAC processor 211 through, amongothers, transmit vector (TXVECTOR) and receive vector (RXVECTOR)parameters. In one or more aspects, the MAC processor 211 generates andprovides TXVECTOR parameters to the PHY processor 215 to supplyper-packet transmit parameters. In one or more aspects, the PHYprocessor 215 generates and provides RXVECTOR parameters to the MACprocessor 211 to inform the MAC processor 211 of the received packetparameters.

In some aspects, the wireless communication device 200 includes aread-only memory (ROM) (not shown) or registers (not shown) that storeinstructions that are needed by one or more of the MAC processor 211,the PHY processor 215 and/or other components of the wirelesscommunication device 200.

In one or more implementations, the wireless communication device 200includes a permanent storage device (not shown) configured as aread-and-write memory device. The permanent storage device may be anon-volatile memory unit that stores instructions even when the wirelesscommunication device 200 is off. The ROM, registers and the permanentstorage device may be part of the baseband processor 210 or be a part ofthe memory 240. Each of the ROM, the permanent storage device, and thememory 240 may be an example of a memory or a computer-readable medium.A memory may be one or more memories.

The memory 240 may be a read-and-write memory, a read-only memory, avolatile memory, a non-volatile memory, or a combination of some or allof the foregoing. The memory 240 may store instructions that one or moreof the MAC processor 211, the PHY processor 215, and/or anothercomponent may need at runtime.

The RF transceiver 220 includes an RF transmitter 221 and an RF receiver222. The input interface unit 250 receives information from a user, andthe output interface unit 260 outputs information to the user. Theantenna unit 230 includes one or more antennas. When multiple-inputmultiple-output (MIMO) or multi-user MIMO (MU-MIMO) is used, the antennaunit 230 may include more than one antenna.

The bus 270 collectively represents all system, peripheral, and chipsetbuses that communicatively connect the numerous internal components ofthe wireless communication device 200. In one or more implementations,the bus 270 communicatively connects the baseband processor 210 with thememory 240. From the memory 240, the baseband processor 210 may retrieveinstructions to execute and data to process in order to execute theprocesses of the subject disclosure. The baseband processor 210 can be asingle processor, multiple processors, or a multi-core processor indifferent implementations. The baseband processor 210, the memory 240,the input interface unit 250, and the output interface unit 260 maycommunicate with each other via the bus 270.

The bus 270 also connects to the input interface unit 250 and the outputinterface unit 260. The input interface unit 250 enables a user tocommunicate information and select commands to the wirelesscommunication device 200. Input devices that may be used with the inputinterface unit 250 may include any acoustic, speech, visual, touch,tactile and/or sensory input device, e.g., a keyboard, a pointingdevice, a microphone, or a touchscreen. The output interface unit 260may enable, for example, the display or output of videos, images, audio,and data generated by the wireless communication device 200. Outputdevices that may be used with the output interface unit 260 may includeany visual, auditory, tactile, and/or sensory output device, e.g.,printers and display devices or any other device for outputtinginformation. One or more implementations may include devices thatfunction as both input and output devices, such as a touchscreen.

One or more implementations can be realized in part or in whole using acomputer-readable medium. In one aspect, a computer-readable mediumincludes one or more media. In one or more aspects, a computer-readablemedium is a tangible computer-readable medium, a computer-readablestorage medium, a non-transitory computer-readable medium, amachine-readable medium, a memory, or some combination of the foregoing(e.g., a tangible computer-readable storage medium, or a non-transitorymachine-readable storage medium). In one aspect, a computer is amachine. In one aspect, a computer-implemented method is amachine-implemented method.

A computer-readable medium may include storage integrated into aprocessor and/or storage external to a processor. A computer-readablemedium may be a volatile, non-volatile, solid state, optical, magnetic,and/or other suitable storage device, e.g., RAM, ROM, PROM, EPROM, aflash, registers, a hard disk, a removable memory, or a remote storagedevice.

In one aspect, a computer-readable medium comprises instructions storedtherein. In one aspect, a computer-readable medium is encoded withinstructions. In one aspect, instructions are executable by one or moreprocessors (e.g., 210, 215, 280, 290) to perform one or more operationsor a method. Instructions may include, for example, programs, routines,subroutines, data, data structures, objects, sequences, commands,operations, modules, applications, and/or functions. Those skilled inthe art would recognize how to implement the instructions.

A processor (e.g., 210, 215, 280, 290) may be coupled to one or morememories (e.g., one or more external memories such as the memory 240,one or more memories internal to the processor, one or more registersinternal or external to the processor, or one or more remote memoriesoutside of the device 200), for example, via one or more wired and/orwireless connections. The coupling may be direct or indirect. In oneaspect, a processor includes one or more processors. A processor,including a processing circuitry capable of executing instructions, mayread, write, or access a computer-readable medium. A processor may be,for example, an application specific integrated circuit (ASIC), adigital signal processor (DSP), or a field programmable gate array(FPGA).

In one aspect, a processor (e.g., 210, 215, 280, 290) is configured tocause one or more operations of the subject disclosure to occur. In oneaspect, a processor is configured to cause an apparatus (e.g., awireless communication device 200) to perform operations or a method ofthe subject disclosure. In one or more implementations, a processorconfiguration involves having a processor coupled to one or morememories. A memory may be internal or external to the processor.Instructions may be in a form of software, hardware or a combinationthereof. Software instructions (including data) may be stored in amemory. Hardware instructions may be part of the hardware circuitrycomponents of a processor. When the instructions are executed orprocessed by one or more processors, (e.g., 210, 215, 280, 290), the oneor more processors cause one or more operations of the subjectdisclosure to occur or cause an apparatus (e.g., a wirelesscommunication device 200) to perform operations or a method of thesubject disclosure.

FIG. 3A illustrates a schematic block diagram of an example of atransmitting signal processing unit in a wireless communication device.The transmitting signal processing unit 280 of the PHY processor 215includes an encoder 281, an interleaver 282, a mapper 283, an inverseFourier transformer (IFT) 284, and a guard interval (GI) inserter 285.

The encoder 281 encodes input data. For example, the encoder 281 may bea forward error correction (FEC) encoder. The FEC encoder may include abinary convolutional code (BCC) encoder followed by a puncturing device,or may include a low-density parity-check (LDPC) encoder. In one or moreimplementations, the transmitting signal processing unit 280 includes ascrambler (not shown) for scrambling the input data before the encodingoperation to reduce the probability of long sequences of zeroes or ones.

If BCC encoding is used in the encoder 281, the transmitting signalprocessing unit 280 may further include an encoder parser fordemultiplexing the scrambled bits among one or more BCC encoders. In oneaspect, if LDPC encoding is used in the encoder 281, the transmittingsignal processing unit 280 may not use the encoder parser.

The interleaver 282 interleaves the bits of each stream output from theencoder 281 to change the order of bits. In one aspect, interleaving maybe applied only when BCC encoding is employed.

The mapper 283 maps the sequence of bits output from the interleaver 282into constellation points. If the LDPC encoding is used in the encoder281, the mapper 283 may further perform LDPC tone mapping instead of theconstellation mapping.

When MIMO or MU-MIMO is employed, the transmitting signal processingunit 280 may use multiple instances of the interleaver 282 and multipleinstances of the mapper 283 corresponding to the number of spatialstreams (Nss). In this example, the transmitting signal processing unit280 may further include a stream parser for dividing outputs of the BCCencoders or the LDPC encoder into blocks that are sent to differentinterleavers 282 or mappers 283. The transmitting signal processing unit280 may further include a space-time block code (STBC) encoder forspreading the constellation points from the number of spatial streamsinto a number of space-time streams (NsTs) and a spatial mapper formapping the space-time streams to transmit chains. The spatial mappermay use direct mapping, spatial expansion, or beamforming depending onimplementation. When MU-MIMO is employed, one or more of the blocksbefore reaching the spatial mapper may be provided for each user.

The IFT 284 converts a block of the constellation points output from themapper 283 or the spatial mapper into a time domain block (e.g., asymbol) by using an inverse discrete Fourier transform (IDFT) or aninverse fast Fourier transform (IFFT). If the STBC encoder and thespatial mapper are employed, the IFT 284 may be provided for eachtransmit chain.

When MIMO or MU-MIMO is employed, the transmitting signal processingunit 280 may insert cyclic shift diversities (CSDs) to preventunintentional beamforming. The CSD insertion may occur before or afterthe inverse Fourier transform operation. The CSD may be specified pertransmit chain or may be specified per space-time stream. Alternatively,the CSD may be applied as a part of the spatial mapper.

The GI inserter 285 prepends a GI to the symbol. The transmitting signalprocessing unit 280 may optionally perform windowing to smooth edges ofeach symbol after inserting the GI. The RF transmitter 221 converts thesymbols into an RF signal and transmits the RF signal via the antennaunit 230. When MIMO or MU-MIMO is employed, the GI inserter 285 and theRF transmitter 221 may be provided for each transmit chain.

FIG. 3B illustrates a schematic block diagram of an example of areceiving signal processing unit in a wireless communication device. Thereceiving signal processing unit 290 of the PHY processor 215 includes aGI remover 291, a Fourier transformer (FT) 292, a demapper 293, adeinterleaver 294, and a decoder 295.

The RF receiver 222 receives an RF signal via the antenna unit 230 andconverts the RF signal into one or more symbols. In some aspects, the GIremover 291 removes the GI from the symbol. When MIMO or MU-MIMO isemployed, the RF receiver 222 and the GI remover 291 may be provided foreach receive chain.

The FT 292 converts the symbol (e.g., the time domain block) into ablock of the constellation points by using a discrete Fourier transform(DFT) or a fast Fourier transform (FFT) depending on implementation. Inone or more implementations, the FT 292 is provided for each receivechain.

When MIMO or MU-MIMO is employed, the receiving signal processing unit290 may be a spatial demapper for converting the Fourier transformedreceiver chains to constellation points of the space-time streams, and aSTBC decoder (not shown) for despreading the constellation points fromthe space-time streams into the spatial streams.

The demapper 293 demaps the constellation points output from the FT 292or the STBC decoder to the bit streams. If the LDPC encoding is used,the demapper 293 may further perform LDPC tone demapping before theconstellation demapping. The deinterleaver 294 deinterleaves the bits ofeach stream output from the demapper 293. In one or moreimplementations, deinterleaving may be applied only when BCC encoding isused.

When MIMO or MU-MIMO is employed, the receiving signal processing unit290 may use multiple instances on the demapper 293 and multipleinstances of the deinterleaver 294 corresponding to the number ofspatial streams. In this example, the receiving signal processing unit290 may further include a stream deparser for combining the streamsoutput from the deinterleavers 294.

The decoder 295 decodes the streams output from the deinterleaver 294and/or the stream deparser. For example, the decoder 295 may be an FECdecoder. The FEC decoder may include a BCC decoder or an LDPC decoder.

The receiving signal processing unit 290 may further include adescrambler for descrambling the decoded data. If BCC decoding is usedin the decoder 295, the receiving signal processing unit 290 may furtherinclude an encoder deparser for multiplexing the data decoded bymultiple instances of the BCC decoder. In one or more implementations,if LDPC decoding is used in the decoder 295, the receiving signalprocessing unit 290 may not use the encoder deparser.

FIG. 4 illustrates an example of a timing diagram of interframe space(IFS) relationships. In this example, a data frame, a control frame, ora management frame can be exchanged between the wireless communicationdevices 111-115 and/or other WLAN devices.

Referring to the timing diagram 400, during the time interval 402,access is deferred while the medium (e.g., a wireless communicationchannel) is busy until a type of IFS duration has elapsed. At timeinterval 404, immediate access is granted when the medium is idle for aduration that is equal to or greater than a distributed coordinationfunction IFS (DIFS) 410 duration or arbitration IFS (AIFS) 414 duration.In turn, a next frame 406 may be transmitted after a type of IFSduration and a contention window 418 have passed. During the time 408,if a DIFS has elapsed since the medium has been idle, a designated slottime 420 is selected and one or more backoff slots 422 are decrementedas long as the medium is idle.

The data frame is used for transmission of data forwarded to a higherlayer. In one or more implementations, a WLAN device transmits the dataframe after performing backoff if DIFS 410 has elapsed from a time whenthe medium has been idle.

The management frame is used for exchanging management information thatis not forwarded to the higher layer. Subtype frames of the managementframe include a beacon frame, an association request/response frame, aprobe request/response frame, and an authentication request/responseframe.

The control frame is used for controlling access to the medium. Subtypeframes of the control frame include a request to send (RTS) frame, aclear to send (CTS) frame, and an ACK frame. In the case that thecontrol frame is not a response frame of the other frame (e.g., aprevious frame), the WLAN device transmits the control frame afterperforming backoff if the DIFS 410 has elapsed. In the case that thecontrol frame is the response frame of the other frame, the WLAN devicetransmits the control frame without performing backoff if a short IFS(SIFS) 412 has elapsed. The type and subtype of frame may be identifiedby a type field and a subtype field in a frame control field of theframe.

On the other hand, a Quality of Service (QoS) STA may transmit the frameafter performing backoff if AIFS 414 for access category (AC), e.g.,AIFS[AC], has elapsed. In this case, the data frame, the managementframe, or the control frame that is not the response frame may use theAIFS[AC].

In one or more implementations, a point coordination function (PCF)enabled AP STA transmits the frame after performing backoff if a PCF IFS(PIFS) 416 has elapsed. In this example, the PIFS 416 duration is lessthan the DIFS 410 but greater than the SIFS 412. In some aspects, thePIFS 416 is determined by incrementing the SIFS 412 duration by adesignated slot time 420.

FIG. 5 illustrates an example of a timing diagram of a carrier sensemultiple access/collision avoidance (CSMA/CA) based frame transmissionprocedure for avoiding collision between frames in a channel. In FIG. 5, any one of the wireless communication devices 111-115 in FIG. 1 can bedesignated as one of STA1, STA2 or STA3. In this example, the wirelesscommunication device 111 is designated as STA1, the wirelesscommunication device 112 is designated as STA2, and the wirelesscommunication device 113 is designated as STA3. While the timing of thewireless communication devices 114 and 115 is not shown in FIG. 5 , thetiming of the devices 114 and 115 may be the same as that of STA2.

In this example, STA1 is a transmit WLAN device for transmitting data,STA2 is a receive WLAN device for receiving the data, and STA3 is a WLANdevice that may be located at an area where a frame transmitted from theSTA1 and/or a frame transmitted from the STA2 can be received by theSTA3.

The STA1 may determine whether the channel (or medium) is busy bycarrier sensing. The STA1 may determine the channel occupation based onan energy level on the channel or correlation of signals in the channel.In one or more implementations, the STA1 determines the channeloccupation by using a network allocation vector (NAV) timer.

When determining that the channel is not used by other devices duringthe DIFS 410 (e.g., the channel is idle), the STA1 may transmit an RTSframe 502 to the STA2 after performing backoff. Upon receiving the RTSframe 502, the STA2 may transmit a CTS frame 506 as a response of theCTS frame 506 after the SIFS 412.

When the STA3 receives the RTS frame 502, the STA3 may set a NAV timerfor a transmission duration representing the propagation delay ofsubsequently transmitted frames by using duration information involvedwith the transmission of the RTS frame 502 (e.g., NAV(RTS) 510). Forexample, the STA3 may set the transmission duration expressed as thesummation of a first instance of the SIFS 412, the CTS frame 506duration, a second instance of the SIFS 412, a data frame 504 duration,a third instance of the SIFS 412 and an ACK frame 508 duration.

Upon receiving a new frame (not shown) before the NAV timer expires, theSTA3 may update the NAV timer by using duration information included inthe new frame. The STA3 does not attempt to access the channel until theNAV timer expires.

When the STA1 receives the CTS frame 506 from the STA2, the STA1 maytransmit the data frame 504 to the STA2 after the SIFS 412 elapses froma time when the CTS frame 506 has been completely received. Uponsuccessfully receiving the data frame 504, the STA2 may transmit the ACKframe 508 after the SIFS 412 elapses as an acknowledgment of receivingthe data frame 504.

When the NAV timer expires, the STA3 may determine whether the channelis busy by the carrier sensing. Upon determining that the channel is notused by the other WLAN devices (e.g., STA1, STA2) during the DIFS 410after the NAV timer has expired, the STA3 may attempt the channel accessafter a contention window 418 has elapsed. In this example, thecontention window 418 may be based on a random backoff.

FIG. 6 illustrates a very high throughput (VHT) frame format 650. TheVHT frame format 650 is a physical layer convergence procedure (PLCP)protocol data unit (or PPDU) format. In the VHT format 650, packets ofthis format contain the legacy preamble 601 composed of a legacy(non-HT) short training field (L-STF), a legacy long training field(L-LTF), and a legacy signal field (L-SIG). The L-STF and L-LTF portionseach has a duration of 8 microsecond (μs) per symbol whereas the L-SIGportion has a duration of 4 μs per symbol. The L-STF field may beutilized for packet detection, automatic gain control (AGC), and coarsefrequency-offset correction. The L-LTF field may be utilized for channelestimation, fine frequency-offset correction, and symbol timing.

The remainder of the VHT frame format 650 includes a VHT specificportion 651 composed of a VHT-SIG-A field, a VHT-STF field, a VHT-LTFfield, and a VHT-SIG-B field. The VHT frame format 650 also includes thedata frame 605, which is composed of a service field, a scrambled PLCPservice data unit (PSDU) field, a tail bits field and a pad bits field.The VHT-SIG-A field has a duration of 8 μs per symbol while the VHT-STF,the VHT-LTF and the VHT-SIG-B fields each has a duration of 4 μs persymbol.

Clause 22 specifies the PHY entity for a very high throughput (VHT)orthogonal frequency-division multiplexing (OFDM) system, which isidentified in the IEEE Standard 802.11ac. In one or moreimplementations, a VHT STA transmits and receives PPDUs that arecompliant with the PHY specifications defined in Clause 20. Clause 20specifies the PHY entity for a HT OFDM system, which is identified inthe IEEE Standard 802.11n. The VHT PHY is based on the HT PHY defined inClause 20, which in turn is based on the OFDM PHY defined in Clause 18.

In one or more implementations, a VHT single-user (SU) PPDU includesindividually addressed and group addressed transmissions. In someaspects, the VHT PHY provides support for 20 MHz, 40 MHz, 80 MHz and 160MHz contiguous channel widths and support for 80+80 MHz non-contiguouschannel width. The VHT PHY data sub-carriers are modulated using binaryphase shift keying (BPSK), quadrature phase shift keying (QPSK),16-quadrature amplitude modulation (16-QAM), 64-QAM and 256-QAM. Forwarderror correction (FEC) coding (convolutional or low-density parity check(LDPC) coding) is used with coding rates of 1/2, 2/3, 3/4 and 5/6.

The VHT-LTF sequences for transmission use channel bandwidths (CBWs) of20 MHz, 40 MHz, 80 MHz, and 160 MHz. Each VHT-LTF sequence is definedover a set of tone indices.

In a 20 MHz transmission, the VHT-LTF sequence transmitted is given byVHTLTF_(−28,28)={1,1, LTF_(left),0, LTF_(right),−1,−1}  Equation (1)whereLTE_(left)={1,1,−1,−1,1,1,−1,1,−1,1,1,1,1,1,1,−1,−1,1,1,−1,1,−1,1,1,1,1}  Equation(2)andLTF_(right)={1,−1,−1,1,1,−1,1,−1,1,−1,−1,−1,−1,−1,1,1,−1,−1,1,−1,1,−1,1,1,1,1}  Equation(3)The VHTLTF_(−28,28) sequence is defined for tone indices from −28 to+28.

In a 40 MHz transmission, the VHT-LTF sequence transmitted is given byVHTLTF_(−58,58)={VHTLTF_(left),0,0,0,VHTLTF_(right)}  Equation (4)whereVHTLTF_(left)={LTF_(left),1,LTF_(right),−1,−1,−1,1}  Equation (5)andVHTLTF_(right)={−1,1,1,−1,LTF_(left),1,LTF_(right)}  Equation (6)The sequence VHTLTF_(−58,58) is defined for tone indices from −58 to+58.

In an 80 MHz transmission, the VHT-LTF sequence transmitted is given byVHTLTF_(−122,122)={LTF_(left),1,LTF_(right),−1,−1,−1,1,1,−1,1,−1,1,1,−1,LTF_(left),1,LTF_(right),1,−1,1,−1,0,0,0,1,−1,−1,1,LTF_(left),1,LTF_(right),−1,−1,−1,1,1,−1,1,−1,1,−1,1,1,−1,LTF_(left),1,LTF_(right)}  Equation(7)or equivalentlyVHTLTF_(−122,122)={VHTLTF_(left),1,−1,1,VHTLTF_(right),1,−1,1,−1,0,0,0,1,−1,−1,1,VHTLTF_(left),1,−1,1,VHTLTF_(right)}  Equation(8)The sequence VHTLTF_(−122,122) is defined for tone indices from −122 to+122.

In a 160 MHz transmission, the VHT-LTF sequence transmitted is given byVHTLTF_(−250,250)={VHTLTF_(−122,122),0,0,0,0,0,0,0,0,0,0,VHTLTF_(−122,122)}  Equation(9)The sequence VHTLTF_(−250,250) is defined for tone indices from −250 to+250.

The time domain representation of a waveform transmitted on frequencysegment i_(Seg) of transmit chain i_(TX) is described by the followingequation:

$\begin{matrix}{{r_{{VHT} - {LTF}}^{({i_{Seg},i_{Tx}})}(t)} = {\frac{1}{\sqrt{N_{{VHT} - {LTF}}^{Tone}N_{{STS},{total}}}}{\sum_{n = 0}^{N_{VHLTF} - 1}{{w_{T_{{VHT} - {LTF}}}\left( {t - {nT}_{{VHT} - {LTF}}} \right)} \cdot {\sum_{k = {- N_{SR}}}^{N_{SR}}{\sum_{u = 0}^{N_{user} - 1}{\sum_{m = 1}^{N_{{STS},u}}\text{ }\begin{pmatrix}{\left\lbrack Q_{k}^{(i_{Seg})} \right\rbrack_{i_{Tx},{({M_{u} + m})}}{\gamma_{k,{BW}}\left\lbrack A_{VHTLTF}^{k} \right\rbrack}_{{({M_{u} + m})},{({n + 1})}}{{VHTLF}_{k} \cdot}} \\{\exp\left( {j2\pi k{\Delta}_{F}\left( {t - {nT}_{{VHT} - {LTF}} - T_{GI} - {T_{{CS},{VHT}}\left( {M_{u} + m} \right)}} \right)} \right)}\end{pmatrix}}}}}}}} & {{Equation}(10)}\end{matrix}$where r_(VHT-LTF) ^((i) ^(Seg) ^(,i) ^(Tx) ⁾ represents the time domainsignal, w_(T) _(VHT-LTF) (t−nT_(VHT-LTF)) represents a time domainwindowing filter, and γ_(k,BW) represents a phase rotation. More detailsare in the IEEE 802.11ac specification. With regard to the summationindices, k represents a tone index, u represents a user index, and mrepresents a spatial stream index (per user).

The γ_(k,BW) is used to represent a rotation of the tones. BW inγ_(k,BW) is determined by the TXVECTOR parameter CH BANDWIDTH asprovided in the following table:

CH_BANDWIDTH and γ_(k,BW) CH_BANDWIDTH γ_(k,BW) CBW20 γ_(k,20) CBW40γ_(k,40) CBW80 γ_(k,80) CBW160 γ_(k,160) CBW80 + 80 γ_(k,80) perfrequency segment

Phase rotation may be applied per 20 MHz bandwidth, such as in IEEE802.11n and 802.11ac.

For a 20 MHz PPDU transmission,γ_(k,20)=1  Equation (11)

For a 40 MHz PPDU transmission,

$\begin{matrix}{\gamma_{k,{40}} = \left\{ \begin{matrix}{1,\ {k < 0}} \\{j,\ {k \geq 0}}\end{matrix} \right.} & {{Equa}ti{on}\ (12)}\end{matrix}$

For an 80 MHz PPDU transmission,

$\begin{matrix}{\gamma_{k,80} = \left\{ \begin{matrix}{1,\ {k < {{- 6}4}}} \\{{- 1},\ {k \geq {{- 6}4}}}\end{matrix} \right.} & {{Equa}ti{on}\ (13)}\end{matrix}$which can be written equivalently as:

$\begin{matrix}{\gamma_{k,80} = \left\{ \begin{matrix}{1,\ {k < {{- 6}4}}} \\{{- 1},\ {{{- 6}4} \leq k < 0}} \\{{- 1},\ {0 \leq k < {64}}} \\{{- 1},\ {{64} \leq k}}\end{matrix} \right.} & {{Equation}(14)}\end{matrix}$

For a 160 MHz PPDU transmission,

$\begin{matrix}{\gamma_{k,160} = \left\{ \begin{matrix}{1,\ {k < {{- 1}92}}} \\{{- 1},\ {{{- 1}92} \leq k < 0}} \\{1,\ {0 \leq k < {64}}} \\{{- 1},\ {{64} \leq k}}\end{matrix} \right.} & {{Equation}(15)}\end{matrix}$

FIGS. 7A through 7D illustrate examples of applying a phase rotation toa VHT-LTF sequence. For each VHT-LTF sequence, the DC tone at tone indexk=0 is labeled. A value above each of the brackets denotes a phaserotation value that will be multiplied (e.g., applied) to the subset ofthe sequence within the bracket.

FIG. 7A illustrates an example of applying the phase rotation providedin Equation (11) to the VHTLTF_(−28,28) sequence provided in Equation(1). In this case, the resulting sequence is the same asVHTLTF_(−28,28), since applying the phase rotation provided by Equation(11) involves multiplying each element of the VHTLTF_(−28,28) sequenceby 1.

FIG. 7B illustrates an example of applying the phase rotation providedin Equation (12) to the VHTLTF_(−58,58) sequence provided in Equation(4). In this case, the resulting sequence is provided by{1·[VHTLTF_(left), 0], j·[0, 0, VHTLTF_(right)] }. The character “·”refers to multiplication.

FIG. 7C illustrates an example of applying the phase rotation providedin Equation (13), or equivalently Equation (14), to theVHTLTF_(−122,122) sequence provided in Equation (8). In this case, theresulting sequence is provided by {1·[VHTLTF_(left), 1], −1·[−1, 1,VHTLTF_(right), 1, −1, 1, −1, 0, 0, 0, 1, −1, −1, 1, VHTLTF_(left), 1,−1, 1, VHTLTF_(right)]}

FIG. 7D illustrates an example of applying the phase rotation providedin Equation (15) to the VHTLTF_(−250,250) sequence provided in Equation(9).

FIG. 8 illustrates an example of a high efficiency (HE) frame 850. TheHE frame 850 is in a PPDU format. The HE frame 850 includes a preambleand a payload. A preamble (e.g., some or all fields before the lastfield in FIG. 8 ) includes a legacy preamble and an HE preamble.

The legacy preamble comprises: L-STF, L-LTF, and L-SIG. Presence ofthese legacy symbols would make any new design compatible with thelegacy designs and products. In one or more implementations, the legacySTF, LTF and SIG symbols are modulated/carried with FFT size of 64 on a20 MHz sub-channel and the modulated symbols are duplicated on every 20MHz sub-channel when a PPDU has a channel bandwidth wider than 20 MHz(e.g., 40 MHz, 80 MHz, 160 MHz, or 80+80 MHz). In addition appropriatephase rotation may be applied.

The HE preamble may include HE SIG-A, HE STF, HE LTF and HE SIG-B. TheHE SIG-A and HE SIG-B are symbols that carry control information thatmay be vital regarding each PSDU and regarding the radio frequency (RF),PHY, and MAC properties of a PPDU. In the present disclosure, severalfields are located either in HE SIG-A and/or HE SIG-B. The HE SIG-A andHE SIG-B can be carried/modulated using FFT size of 64 or 256 dependingon implementation. In some implementations, an HE SIG-B is not presentin a PPDU.

The HE short training field (STF) and HE long training field (LTF) aresymbols used to perform necessary RF and PHY processing for each PSDUand/or for the whole PPDU. Depending on whether the HE STF/LTF symbolsare beamformed, there may be two sets of such symbols. In one or moreimplementations, the present disclosure constructs LTF sequences thatare utilized in an HE LTF field of an HE frame 850. In one or moreaspects, an HE LTF sequence is predetermined. In one or more aspects, anHE LTF sequence is not random data. In one or more aspect, an HE LTFsequence is predetermined for each channel bandwidth and/or for eachsubband utilized. In one or more aspects, an HE LTF sequence ispredetermined or selected based on any one or more of the following: oneor more other HE LTF sequences, one or more LTF sub-sequences, one ormore phase rotation values, and/or one or more subband allocation. Inone or more aspects, an HE LTF sequence may be determined, selected,obtained, generated, or provided based on preexisting sequence,sub-sequence, information, functions, formula and/or method.

A payload (e.g., the last field depicted as data in FIG. 8 ) may includeone or more payloads, or one or more data frames. A data frame mayinclude a PLCP service data unit (PSDU), which includes data. In one ormore aspects, data in a payload can include audio, image, video, textand/or other type of data. Such data (e.g., a video sequence) can changeover time. In one aspect, such data is not predetermined. Data may beoutputted at a STA using a visual, auditory, tactile, and/or sensoryoutput device.

An HE frame may be a downlink (DL) frame or an uplink (UL) frame. An HEframe for transmission is associated with one of the channel bandwidths(e.g., 20 MHz, 40 MHz, 80 MHz, 160 MHz, or 80+80 MHz). In other words,an HE frame has a channel bandwidth, and it utilizes (or occupies) itschannel bandwidth. An 80+80 MHz channel bandwidth utilizes (or occupies)two 80 MHz channel bandwidths, where the first 80 MHz bandwidth and thesecond 80 MHz bandwidth are non-contiguous.

In one aspect, a downlink frame may refer to a DL OFDMA frame, an HE DLOFDMA frame, a DL OFDMA PPDU, an HE DL OFDMA PPDU, a DL PPDU, or viceversa. In one aspect, an uplink frame may refer to a UL OFDMA frame, anHE UL OFDMA frame, a UL OFDMA PPDU, an HE UL OFDMA PPDU, a UL PPDU, orvice versa. In one aspect, a PPDU may refer to an HE PPDU or an OFDMAPPDU. In one aspect, a PPDU is a downlink frame or an uplink frame.

In one aspect, a preamble is referred to as a preamble header, apreamble field, a preamble section, a header, a header section, or viceversa. For the sake of brevity, a preamble may refer to a component of apreamble. Thus, in one aspect, a preamble may refer to one or morepreambles (e.g., L-STF, L-LTF, L-SIG, HE SIG-A, HE STF, HE LTF, and/orHE SIG-B).

In one aspect, a legacy preamble (e.g., each or all of L-STF, L-LTF andL-SIG) is associated with a channel bandwidth of a frame (e.g., downlinkframe or uplink frame). In one example, for a given channel bandwidth(e.g., 80 MHz) of a frame, a legacy preamble is modulated on the entirechannel bandwidth of the frame (e.g., entire 80 MHz). In anotherexample, a legacy preamble is modulated on a sub-channel (e.g., 20 MHzsub-channel) of the channel bandwidth (e.g., 80 MHz) and the modulatedsignal is duplicated on each of the remaining sub-channels (e.g.,remaining three 20 MHz sub-channels) of the channel bandwidth to allowthe legacy preamble to utilize the entire channel bandwidth.

In one aspect, an access point may allocate different portions of achannel bandwidth to different STAs. In one aspect, a portion of achannel bandwidth is allocated to a STA, and an HE LTF for the STA isassociated with the portion of the channel bandwidth. In one aspect, aportion of a channel bandwidth may be a resource unit. In anotheraspect, a portion of a channel bandwidth may be one or more resourceunits. In yet another aspect, a portion of a channel bandwidth may beone or more blocks of a channel bandwidth.

One or more aspects of the present disclosure provide techniques forconstructing an HE LTF sequence for an HE preamble of an HE-basedtransmission. In one aspect, an HE LTF sequence may be utilized forchannel estimation. Channel estimation may be utilized to decode datatransmitted and compensate for channel properties (e.g., effects,distortions). For example, when a preamble is transmitted through awireless channel, various distortions may occur, and a training sequencein the HE LTF field is useful to reverse the distortion. This may bereferred to as equalization. To accomplish this, the amount of channeldistortion is measured. This may be referred to as channel estimation.In one aspect, channel estimation is performed using an HE LTF sequence,and the channel estimation may be applied to other fields that followthe HE LTF sequence.

Examples of numerologies are provided below for each of the 20 MHz, 40MHz, 80 MHz, 160 MHz, and 80+80 MHz HE channel bandwidths. A channelbandwidth may be sometimes referred to as a transmission bandwidth. Inthe description provided below, for the sake of brevity, in one or moreaspects, an LTF may refer to an HE LTF, an LTF sequence may refer to anHE LTF sequence, a sequence may refer to an HE LTF sequence, and a phaserotation block may refer to an HE phase rotation block.

FIG. 9 illustrates an example numerology for a 20 MHz channel bandwidth.The numerology provides different manners by which to allocate resourcesfor the 20 MHz channel bandwidth into individual resource units. Aresource unit contains tones, where each tone may be a data tone or apilot tone.

A tone may be referred to as subcarrier. Each tone may be associatedwith or otherwise identified by a tone index or a subcarrier index. Atone index may be referred to as a subcarrier index.

For a 20 MHz HE PPDU transmission, the 20 MHz may be divided into 256tones with a 78.125 kHz spacing between tones. In this case, a signalmay be transmitted on tone indices−122 to −2 and 2 to 122, with the toneindices−1, 0, and +1 being direct current (DC) tones. Hence, there maybe a total of 242 usable tones, which do not include DC tones. Theremaining 11 tones may be guard tones, where 6 tones may be for one edgeof the bandwidth, and 5 tones may be for the other edge of thebandwidth. (See 901 and 902 in FIG. 9 ). In one aspect, usable tones donot include DC tones or guard tones.

A first row 905 illustrates an example of usable tones for a 20 MHzchannel bandwidth. In one aspect, usable tones include data/pilot tonesand any reserved tones. A data/pilot tone is a data tone or a pilottone. A data/pilot tone is a tone that can be utilized as a data tone ora pilot tone. A reserved tone may be referred to as a null tone or aleft over tone. A reserved tone may have zero energy.

A second row 910 illustrates a resource allocation of the 20 MHzbandwidth into multiple resource units. In one or more implementations,the 20 MHz bandwidth may be allocated into 9 resource units. Eachnon-center resource unit includes 26 data/pilot tones. A center resourceunit includes 26 data/pilot tones and 3 DC tones. A center resource unitmay be referred to as a central resource unit. One reserved tone may beallocated between two adjacent resource units. For example, theallocation may include, in order from the lowest usable tone index(e.g., −122) to the highest usable tone index (e.g., +122) plus DCtones: 26 data/pilot tones, 1 reserved tone, 26 data/pilot tones, 1reserved tone, 26 data/pilot tones, 1 reserved tone, 26 data/pilottones, 1 reserved tone, 13 data/pilot tones, 3 DC tones, 13 data/pilottones, 1 reserved tone, 26 data/pilot tones, 1 reserved tone, 26data/pilot tones, 1 reserved tone, 26 data/pilot tones, 1 reserved tone,and 26 data/pilot tones. For simplicity, this may be illustrated by thefollowing shorthand resource allocation representation:26-1-26-1-26-1-26-1-13-3-13-1-26-1-26-1-26-1-26. In this example, each“−” is inserted simply for convenience, and it may be used to separatedifferent tone types (e.g., data/pilot, reserved, and DC tones). Each“26” represents 26 data/pilot tones, each “1” represents one reservedtone, each “13” represents 13 data/pilot tones, and the “3” represents 3DC tones. Similar conventions may be used for other shorthand resourceallocation representations provided below. For each 26-tone resourceunit, 24 of the data/pilot tones may be data tones and 2 may be pilottones.

A third row 915 illustrates a resource allocation of the 20 MHzbandwidth into fewer, but generally larger, resource units than theresource allocation illustrated in the second row 910. In one or moreimplementations, the 20 MHz bandwidth may be allocated into 5 resourceunits in the following manner: 4 resource units (each including 52data/pilot tones) and one center resource unit (including 26 data/pilottones and 3 DC tones). Two consecutive reserved tones may be allocatedbetween any two adjacent resource units. For example, the allocation mayinclude, in order from the lowest usable tone index to the highestusable tone index plus DC tones: 52 data/pilot tones, 2 reserved tones,52 data/pilot tones, 2 reserved tones, 13 data/pilot tones, 3 DC tones,13 data/pilot tones, 2 reserved tones, 52 data/pilot tones, 2 reservedtones, and 52 data/pilot tones. For simplicity, this may be illustratedby the following shorthand resource allocation representation:52-2-52-2-13-3-13-2-52-2-52. This example uses conventions similar tothose describe above. Each “52” represents 52 data/pilot tones, and each“2” represents 2 reserved tone. For each 52-tone resource unit, 48 ofthe data/pilot tones may be data tones and 4 may be pilot tones.

A fourth row 920 illustrates a resource allocation of the 20 MHzbandwidth into fewer, but generally larger, resource units than theresource allocation illustrated in the third row 915. In one or moreimplementations, the 20 MHz bandwidth may be allocated into 3 resourceunits in the following manner: 2 resource units (each including 106data/pilot tones) and one center resource unit (including 26 data/pilottones and 3 DC tones). Two consecutive reserved tones may be allocatedbetween any two adjacent resource units. For example, the allocation mayinclude, in order from the lowest usable tone index to the highestusable tone index plus DC tones: 106 data/pilot tones, 2 reserved tones,13 data/pilot tones, 3 DC tones, 13 data/pilot tones, 2 reserved tones,and 106 data/pilot tones. For simplicity, this may be illustrated by thefollowing shorthand resource allocation representation:106-2-13-3-13-2-106. This example uses conventions similar to thosedescribed above. Each “106” represents 106 data/pilot tones, and each“2” represents 2 reserved tone. For each 106-tone resource unit, 102 ofthe data/pilot tones may be data tones and 4 may be pilot tones.

Alternatively, in one or more implementations, the 20 MHz bandwidth maybe allocated into 3 resource units in the following manner: 2 resourceunits (each including 108 data/pilot tones) and one center resource unit(including 26 data/pilot tones and 3 DC tones). In such implementations,no reserved tones are utilized. For example, the allocation may include,in order from the lowest usable tone index to the highest usable toneindex plus DC tones: 108 data/pilot tones, 13 data/pilot tones, 3 DCtones, 13 data/pilot tones, and 108 data/pilot tones. For simplicity,this may be illustrated by the following shorthand resource allocationrepresentation: 108-13-3-13-108. This example uses conventions similarto those describe above. Each “108” represents 108 data/pilot tones. Foreach 108-tone resource unit, 102 of the data/pilot tones may be datatones and 6 may be pilot tones.

For each of the resource allocations illustrated in the second row 910,third row 915, and fourth row 920, one station (STA) may be allocated toone or more of the resource units. For example, for the fourth row 920,a first STA may be allocated to the leftmost resource unit containing108 data/pilot tones, a second STA may be allocated to the centerresource unit containing 26 data/pilot tones and 3 DC tones, and/or athird STA may be allocated to the rightmost resource unit containing 108data/pilot tones. In one or more implementations, a STA that isallocated to the center resource unit may not be allocated to any of theother resource units.

A fifth row 925 illustrates a resource allocation of the 20 MHzbandwidth into a single 242 tone resource unit. For example, theresource allocation may include, in order from the lowest usable toneindex to the highest usable tone index plus DC tones: 121 data/pilottones, 3 DC tones, and 121 data/pilot tones. The allocation may beutilized for a non-OFDMA case, in which all the data/pilot tones (e.g.,the 242 data/pilot tones) are allocated to a single STA.

FIG. 10 illustrates an example numerology for a 20 MHz channelbandwidth. Compared to the example resource allocation illustrated inFIG. 9 , the position (e.g., tone) of the null positions are different.A first row 1005 is the same as the first row 905.

A second row 1010 illustrates a resource allocation of the 20 MHzbandwidth into multiple resource units. In one or more implementations,the 20 MHz bandwidth may be allocated into 9 resource units. Eachnon-center resource unit includes 26 data/pilot tones. A center resourceunit includes 26 data/pilot tones and 3 DC tones. Two reserved tones maybe allocated after every two 26-tone resource units for the negativetone index region when observed from the lowest usable tone index (e.g.,−122), and two reserved tones may be allocated after every two 26-toneresource units for the positive tone index region when observed from thehighest usable tone index (e.g., +122). For example, the resourceallocation may include, in order from the lowest usable tone index(e.g., −122) to the highest usable tone index (e.g., +122) plus DCtones: 26 data/pilot tones, 26 data/pilot tones, 2 reserved tones, 26data/pilot tones, 26 data/pilot tones, 2 reserved tones, 13 data/pilottones, 3 DC tones, 13 data/pilot tones, 2 reserved tones, 26 data/pilottones, 26 data/pilot tones, 2 reserved tones, 26 data/pilot tones, and26 data/pilot tones. For simplicity, this may be illustrated by thefollowing shorthand resource allocation representation:26-26-2-26-26-2-13-3-13-2-26-26-2-26-26. This example uses conventionssimilar to those described above. Each “26” represents 26 data/pilottones, each “2” represents two reserved tones, each “13” represents 13data/pilot tones, and the “3” represents 3 DC tones.

A third row 1015, fourth row 1020, and fifth row 1025 illustrates aresource allocation of the 20 MHz bandwidth into fewer, but generallylarger, resource units than the resource allocation illustrated in thesecond row 1010. The third row 1015, fourth row 1020, and fifth row 1025may be the same as the third row 915, fourth row 920, and fifth row 925of FIG. 9 , in which case the same description applies.

Similar to FIG. 9 , in the second row 1010, third row 1015, and fourthrow 1020 of FIG. 10 , one STA may be allocated to one or more of theresource units. Furthermore, in one or more implementations, a STA thatis allocated to the center resource unit may not be allocated to any ofthe other resource units. The resource allocation illustrated in thefifth row 1025 of FIG. 10 may be utilized for a non-OFDMA case, in whichall the data/pilot tones (e.g., the 242 data/pilot tones) are allocatedto a single STA.

The numerologies for 20 MHz as illustrated in FIGS. 9 and 10 are similarto 80 MHz for IEEE 802.11ac. Whereas IEEE 802.11ac utilizes 64 FFT, HEtechnology may utilize 256 FFT in one or more implementations. In oneaspect, for a given FFT size, the number of tones is given; however,depending on the tone spacing, two OFDM symbols with e.g., FFT=64 andFFT=256 may occupy the same bandwidth.

FIG. 11 illustrates an example numerology for a 40 MHz channelbandwidth. For a 40 MHz HE PPDU transmission, the 40 MHz may be dividedinto 512 tones with a 78.125 kHz spacing between tones. A signal istransmitted on tone indices−244 to −3 and +3 to +244, with the toneindices between −2 and +2, inclusive, being DC tones (i.e., 5 DC tones).Hence, there may be a total of 484 usable tones (not including 5 DCtones). The remaining tones may be guard tones (e.g., 12 guard tones onthe left edge and 11 guard tones on the right edge of the bandwidth). Afirst row 1105 illustrates an example of usable tones spanning toneindices−244 to +244. A usable tone may be utilized as a data/pilot toneor a reserved tone. A reserved tone may be referred to as a null tone.

A second row 1110 illustrates a resource allocation of the 40 MHzbandwidth into multiple resource units. In one or more implementations,the 40 MHz bandwidth may be allocated into 18 resource units thatinclude 26 data/pilot tones each. One reserved tone may be allocatedbetween two adjacent resource units, excluding the DC tone regions. Forexample, the resource allocation may include, in order from the lowestusable tone index (e.g., −244) to the highest usable tone index (e.g.,+244) plus DC tones: 26 data/pilot tones, 1 reserved tone, 26 data/pilottones, 1 reserved tone, 26 data/pilot tones, 1 reserved tone, 26data/pilot tones, 1 reserved tone, 26 data/pilot tones, 1 reserved tone,26 data/pilot tones, 1 reserved tone, 26 data/pilot tones, 1 reservedtone, 26 data/pilot tones, 1 reserved tone, 26 data/pilot tones, 5 DCtones, 26 data/pilot tones, 1 reserved tone, 26 data/pilot tones, 1reserved tone, 26 data/pilot tones, 1 reserved tone, 26 data/pilottones, 1 reserved tone, 26 data/pilot tones, 1 reserved tone, 26data/pilot tones, 1 reserved tone, 26 data/pilot tones, 1 reserved tone,26 data/pilot tones, 1 reserved tone, and 26 data/pilot tones. Forsimplicity, this may be illustrated by the following shorthand resourceallocation representation:26-1-26-1-26-1-26-1-26-1-26-1-26-1-26-1-26-5-26-1-26-1-26-1-26-1-26-1-26-1-26-1-26-1-26.This example uses conventions similar to those described above. Each“26” represents 26 data/pilot tones, each “1” represents one reservedtone, and the “5” represents 5 DC tones.

A third row 1115 illustrates a resource allocation of the 40 MHzbandwidth into fewer, but generally larger, resource units than theresource allocation illustrated in the second row 1110. In one or moreimplementations, the 40 MHz bandwidth may be allocated into a total of10 resource units, having (a) 8 resource units that include 52data/pilot tones each and (b) 2 resource units that include 26data/pilot tones each. Two consecutive reserved tones may be allocatedbetween two adjacent resource units, excluding the DC tone regions. Forexample, the resource allocation may include, in order from the lowestusable tone index to the highest usable tone index plus DC tones: 52data/pilot tones, 2 reserved tones, 52 data/pilot tones, 2 reservedtones, 26 data/pilot tones, 2 reserved tones, 52 data/pilot tones, 2reserved tones, 52 data/pilot tones, 5 DC tones, 52 data/pilot tones, 2reserved tones, 52 data/pilot tones, 2 reserved tones, 26 data/pilottones, 2 reserved tones, 52 data/pilot tones, 2 reserved tones, and 52data/pilot tones. For simplicity, this may be illustrated by thefollowing shorthand resource allocation representation:52-2-52-2-26-2-52-2-52-5-52-2-52-2-26-2-52-2-52. This example usesconventions similar to those described above. Each “52” represents 52data/pilot tones, each “2” represents 2 reserved tones, and the “5”represents 5 DC tones.

A fourth row 1120 illustrates a resource allocation of the 40 MHzbandwidth into fewer, but generally larger, resource units than theresource allocation illustrated in the third row 1115. In one or moreimplementations, the 40 MHz bandwidth may be allocated into a total of 6resource units, having (a) 4 resource units that include 106 data/pilottones each and (b) 2 resource units that include 26 data/pilot toneseach. Two consecutive reserved tones may be allocated between any twoadjacent resource units, excluding the DC tone regions. For example, theresource allocation may include, in order from the lowest usable toneindex to the highest usable tone index plus DC tones: 106 data/pilottones, 2 reserved tones, 26 data/pilot tones, 2 reserved tones, 106data/pilot tones, 5 DC tones, 106 data/pilot tones, 2 reserved tones, 26data/pilot tones, 2 reserved tones, and 106 data/pilot tones. Forsimplicity, this may be illustrated by the following shorthand resourceallocation representation: 106-2-26-2-106-5-106-2-26-2-106. This exampleuses conventions similar to those describe above. Each “106” represents106 data/pilot tones, each “2” represents 2 reserved tones, and the “5”represents 5 DC tones.

Alternatively, in one or more implementations, the 40 MHz bandwidth maybe allocated into a total of 6 resource units that include (a) 4resource units that include 108 data/pilot tones each and (b) 2 resourceunits that include 26 data/pilot tones each. In such implementations, noreserved tones are utilized. For example, the allocation may include, inorder from the lowest usable tone index to the highest usable tone indexplus DC tones: 108 data/pilot tones, 26 data/pilot tones, 108 data/pilottones, 5 DC tones, 108 data/pilot tones, 26 data/pilot tones, and 108data/pilot tones. For simplicity, this may be illustrated by thefollowing shorthand resource allocation representation:108-26-108-5-108-26-108. This example uses conventions similar to thosedescribed above. Each “108” represents 108 data/pilot tones, each “26”represents 26 data/pilot tones, and the “5” represents 5 DC tones.

In one or more implementations, the resource allocation for 40 MHzchannel bandwidth can be obtained through duplications (e.g., two) ofthe resource allocation for 20 MHz channel bandwidth. For example, eachof the left half and the right half of the resource allocation in thesecond row 1110, third row 1115, and fourth row 1120 of FIG. 11 may bethe resource unit allocation illustrated in the second row 910, thirdrow 915, and fourth row 920, respectively, of FIG. 9 , except with theDC tones removed. When constructing the 40 MHz resource allocation basedon duplications (e.g., two) of the 20 MHz resource allocation, each ofthe 26-tone center resource units illustrated in the second throughfourth rows of FIG. 9 (excluding the DC tones) becomes the respectiveone of the 26-tone resource units illustrated in the second throughfourth rows of FIG. 11 on each of the left half and the right half.

A fifth row 1125 illustrates a resource allocation of the 40 MHzbandwidth into fewer, but generally larger, resource units than theresource allocation illustrated in the fourth row 1120. In one or moreimplementations, the 40 MHz bandwidth may be allocated into 2 resourceunits that include 242 data/pilot tones each. In such implementations,no reserved tones are utilized. For example, the allocation may include,in order from the lowest usable tone index to the highest usable toneindex plus DC tones: 242 data/pilot tones, 5 DC tones, and 242data/pilot tones. For simplicity, this may be illustrated by thefollowing shorthand resource allocation representation: 242-5-242. Thisexample uses conventions similar to those describe above. Each “242”represents 242 data/pilot tones, and the “5” represents 5 DC tones. Foreach 242-tone resource unit, 234 of the data/pilot tones may be datatones and 8 may be pilot tones.

For each of the allocations illustrated in the second through fifth rowsof FIG. 11 , one STA may be allocated to one or more of the resourceunits. For example, for the fifth row 1125, one STA may be allocated toone of the 242-tone resource unit, and another STA may be allocated tothe other of 242-tone resource unit.

A sixth row 1130 illustrates a resource allocation of the 40 MHzbandwidth into a single resource unit with 484 non-DC tones. Forexample, the single resource unit may include, in order from the lowestusable tone index to the highest usable tone index plus DC tones: 242data/pilot tones, 5 DC tones, and 242 data/pilot tones. The singleresource unit may be utilized for a non-OFDMA case, in which all thedata/pilot tones are allocated to a single STA.

FIG. 12 illustrates an example numerology for an 80 MHz channelbandwidth. For an 80 MHz HE PPDU transmission, the 80 MHz may be dividedinto 1024 tones with a 78.125 kHz spacing between tones. In one or moreimplementations, the number of usable tones for the OFDMA case may bedifferent from the number of usable tones for the non-OFDMA case. Insome aspects, the number of DC tones for the OFDMA case may be differentfrom the number of DC tones for the non-OFDMA case.

For an 80 MHz HE PPDU transmission for the OFDMA case, a signal may betransmitted on tones −500 to −4 and +4 to +500, with the tones between−3 and +3, inclusive, being DC tones (i.e., 7 DC tones). In thisexample, the number of usable tones may be 994, not including the DCtones. The remaining tones may be guard tones (e.g., 12 guard tones onthe left edge and 11 guard tones on the right edge).

For an 80 MHz HE PPDU transmission for the non-OFDMA case, a signal maybe transmitted on tones −500 to −3 and +3 to +500, with the tonesbetween −2 and +2, inclusive, being DC tones (i.e., 5 DC tones). In thisexample, the number of usable tones may be 996 (excluding the DC tones).The remaining tones may be guard tones (e.g., 12 guard tones on the leftedge and 11 guard tones on the right edge).

In one or more implementations, the resource allocation for 80 MHzchannel bandwidth can be obtained through multiple (e.g., four)duplications of the resource allocation for 20 MHz channel bandwidth orthrough duplications (e.g., two) of the resource allocation for 40 MHzchannel bandwidth.

A first row 1205 illustrates usable tones. The usable tones may spantone indices−500 to +500 for the OFDMA case (excluding −3 to +3 for 7 DCtones) and tone indices−500 to +500 for the non-OFDMA case (excluding −2to +2 for 5 DC tones). A usable tone may be utilized as a data/pilottone or a reserved tone. A reserved tone may be referred to as a nulltone.

A second row 1210 illustrates a resource allocation of the 80 MHzbandwidth into multiple resource units. In one or more implementations,the 80 MHz bandwidth may be allocated into a total of 37 resource units,having (a) 36 resource units that include 26 data/pilot tones each and(b) a center resource unit that includes 26 data/pilot tones and 7 DCtones. One reserved tone may be allocated between any two adjacentresource units, except between the center resource unit and the 26-toneresource units adjacent to the center resource unit. In one or moreimplementations, a STA that is allocated to the center resource unit maynot be allocated to any of the other resource units.

An example of a resource allocation may be illustrated by the followingshorthand resource allocation representation, in order from the lowestusable tone index (e.g., −500) to the highest usable tone index (+500)plus DC tones:

26-1-26-1-26-1-26-1-26-1-26-1-26-1-26-1-26-26-1-26-1-26-1-26-1-26-1-26-1-26-1-26-1-26-13-7-13-26-1-26-1-26-1-26-1-26-1-26-1-26-1-26-1-26-26-1-26-1-26-1-26-1-26-1-26-1-26-1-26-1-26.This example uses conventions similar to those described above. Each“26” represents 26 data/pilot tones, each “1” represents 1 reservedtone, each “13” represents 13 data/pilot tones, and the “7” represents 7DC tones.

A third row 1215 illustrates a resource allocation of the 80 MHzbandwidth into fewer, but generally larger, resource units than theresource allocation illustrated in the second row 1210. In one or moreimplementations, the 80 MHz bandwidth may be allocated into a total of21 resource units, having (a) 16 resource units that include 52data/pilot tones each, (b) 4 resource units that include 26 data/pilottones each, and (c) a center resource unit that includes 26 data/pilottones and 7 DC tones. Two consecutive reserved tones may be allocatedbetween at least some of two adjacent resource units. An example of aresource allocation may be illustrated by the following shorthandresource allocation representation, in order from the lowest usable toneindex to the highest usable tone index plus DC tones:52-2-52-2-26-2-52-2-52-52-2-52-2-26-2-52-2-52-13-7-13-52-2-52-2-26-2-52-2-52-52-2-52-2-26-2-52-2-52. This example uses conventions similar to thosedescribed above. Each “52” represents 52 data/pilot tones, each “26”represents 26 data/pilot tones, each “2” represents 2 reserved tones,each “13” represents 13 data/pilot tones, and the “7” represents 7 DCtones.

A fourth row 1220 illustrates a resource allocation of the 80 MHzbandwidth into fewer, but generally larger, resource units than theresource allocation illustrated in the third row 1215. In one or moreimplementations, the 80 MHz bandwidth may be allocated into a total of13 resource units, having (a) 8 resource units that include 106data/pilot tones each, (b) 4 resource units that include 26 data/pilottones each, and (c) a center resource unit that includes 26 data/pilottones and 7 DC tones. Two consecutive reserved tones may be allocatedbetween some of two adjacent resource units, in a manner that duplicatesthe fourth row 1120 (excluding the DC tones). An example of a resourceallocation may be illustrated by the following shorthand resourceallocation representation, in order from the lowest usable tone index tothe highest usable tone index plus DC tones:106-2-26-2-106-106-2-26-2-106-13-7-13-106-2-26-2-106-106-2-26-2-106.This example uses conventions similar to those described above. Each“106” represents 206 data/pilot tones, each “26” represents 26data/pilot tones, each “2” represents 2 reserved tones, each “13”represents 13 data/pilot tones, and the “7” represents 7 DC tones.

Alternatively, in one or more implementations, the 80 MHz bandwidth maybe allocated into a total of 13 resource units, having (a) 8 resourceunits that include 108 data/pilot tones each, (b) 4 resource units thatinclude 26 data/pilot tones each, and (c) a center resource unit thatincludes 26 data/pilot tones and 7 DC tones. In such implementations, noreserved tones are utilized. An example of a resource allocation may beillustrated by the following shorthand resource allocationrepresentation, in order from the lowest usable tone index to thehighest usable tone index plus DC tones:108-26-108-108-26-108-13-7-13-108-26-108-108-26-108. This example usesconventions similar to those described above. Each “108” represents 108data/pilot tones, each “26” represents 26 data/pilot tones, each “13”represents 13 data/pilot tones, and the “7” represents 7 DC tones.

A fifth row 1225 illustrates a resource allocation of the 80 MHzbandwidth into fewer, but generally larger, resource units than theresource allocation illustrated in the fourth row 1220. In one or moreimplementations, the 80 MHz bandwidth may be allocated into a total of 5resource units, having (a) 4 resource units that include 242 data/pilottones each and (b) a center resource unit that includes 26 data/pilottones and 7 DC tones. In such implementations, no reserved tones areutilized. An example of a resource allocation may be illustrated by thefollowing shorthand resource allocation representation, in order fromthe lowest usable tone index to the highest usable tone index plus DCtones: 242-242-13-7-13-242-242. This example uses conventions similar tothose described above. Each “242” represents 242 data/pilot tones, each“13” represents 13 data/pilot tones, and the “7” represents 7 DC tones.

A sixth row 1230 illustrates a resource allocation of the 80 MHzbandwidth into fewer, but generally larger, resource units than theresource allocation illustrated in the fifth row 1225. In one or moreimplementations, the 80 MHz bandwidth may be allocated into a total of 3resource units, having (a) 2 resource units that include 484 data/pilottones each and (b) a center resource unit that includes 26 data/pilottones and 7 DC tones. In such implementations, no reserved tones areutilized. An example of a resource allocation may be illustrated by thefollowing shorthand resource allocation representation, in order fromthe lowest usable tone index to the highest usable tone index plus DCtones: 484-13-7-13-484. This example uses conventions similar to thosedescribed above. Each “484” represents 484 data/pilot tones, each “13”represents 13 data/pilot tones, and the “7” represents 7 DC tones.

A seventh row 1235 illustrates a resource allocation of the 80 MHzbandwidth into a single resource unit (e.g., non-OFDMA case). In someaspects, the non-OFDMA case utilizes 5 DC tones, and thus the singleresource unit contains 996 non-DC tones.

In one or more implementations, a resource allocation for a 160 MHzchannel bandwidth can be obtained through multiple (e.g., eight)duplications of the resource allocation for 20 MHz channel bandwidth,through multiple (e.g., four) duplications of the resource allocationfor 40 MHz channel bandwidth, or through duplications (e.g., two) of theresource allocation for 80 MHz channel bandwidth.

In one or more implementations, a resource allocation for 80+80 MHzchannel bandwidths can be obtained through multiple (e.g., eight)duplications of the resource allocation for 20 MHz channel bandwidth,through multiple (e.g., four) duplications of the resource allocationfor 40 MHz channel bandwidth, or through duplications (e.g., two) of theresource allocation for 80 MHz channel bandwidth.

Examples of the number of maximum tones, the number of guard tones, thenumber of DC tones, the number of reserved tones, the number of pilottones, and the number of data tones for each of the channel bandwidthsare provided in the table below for a give spacing of 78.125 kHz. Thetable is provided as a non-limiting example.

No. of Used Tones (includes No. of No. of No. of reserved tones, ChannelMaximum Guard DC pilot tones, Bandwidth Spacing Tones Tones Tones anddata tones) 20 MHz 78.125 kHz 256 11 3 234 (non-OFDMA) 20 MHz 78.125 kHz256 11 7 238 (OFDMA) 40 MHz 78.125 kHz 512 23 5 484 (non-OFDMA) 40 MHz78.125 kHz 512 23 5 484 (OFDMA) 80 MHz 78.125 kHz 1024 23 7 994(non-OFDMA) 80 MHz 78.125 kHz 1024 23 5 996 (OFDMA) 160 MHz 78.125 kHz2048 46 14 1988 (non-OFDMA) 160 MHz 78.125 kHz 2048 46 10 1992 (OFDMA)

For each of the 20 MHz, 40 MHz, 80 MHz, 160 MHz, and 80+80 MHz HEchannel bandwidths, various phrase rotation blocks may be utilized.Although examples of phase rotation block options are provided below,other phase rotation block options are possible and may be utilizedinstead.

A rotation of tones may be referred to as a phase rotation, a phaserotation operation, a tone rotation, a tone rotation operation, asubcarrier rotation, or a subcarrier rotation operation. In one aspect,a phase rotation value may be a real or complex value (e.g., +1, −1, j,−j). In one aspect, a phase rotation operation involves multiplying areal or complex value to the LTF sequence.

In some aspects, a phase rotation operation involves multiplying asubset of elements of the LTF sequence that are consecutive in thefrequency domain. The subset of elements is associated with a set oftone indices or, equivalently, a frequency band corresponding to the setof tone indices. The set of tone indices over which a phase rotationoperation is applied may be referred to as a block, a phase rotationblock, a tone rotation block, or a subcarrier rotation block. In one ormore implementations, consecutive phase rotation blocks that have thesame phase rotation value may be consolidated into a single phaserotation block associated with the same phase rotation value.

Phase rotation may be applied to training sequences (e.g., HE LTFsequence) to reduce peak-to-average power ratio (PAPR). In one aspect,values utilized for phase rotation operations may be selected to allowfor a reduction of the PAPR of the LTF OFDM symbols and/or thetransmitted signal. In one or more implementations, a time domainrepresentation of a waveform transmitted on frequency segment i_(Seg) oftransmit chain i_(Tx) may be, for example, similar to Equation (10)provided above, except that VHT is replaced with HE.

Examples of phase rotation blocks are provided as follows for a 20 MHzchannel bandwidth.

For a 20 MHz channel bandwidth, a phase rotation may be applied in unitsof 5 MHz, corresponding to 64 tones. The phase rotation may bemultiplied for each consecutive 64 tones, including guard tones.Accordingly, the four phase rotation blocks may be provided such thatphase rotation block 1 includes tone indices {−122, −65}, phase rotationblock 2 includes tone indices {−64, −1}, phase rotation block 3 includestone indices {1, 64}, and phase rotation block 4 includes tone indices{65, 122} for 20 MHz bandwidth, where the notation {a,b} represents toneindices a to b, inclusive. The phase rotation values may be given by[c₁, c₂, c₃, c₄]. In this example, elements with tone indices−122 to −65are multiplied by c₁, elements with tone indices−64 to −1 are multipliedby c₂, and so on. For example, the LTF field may utilizeVHTLTF_(−122,122) as the sequence and phase rotation values [+1, −1, −1,−1].

In one or more implementations, due to the modular resource allocationin OFDMA, the phase rotation blocks may be designed based on the OFDMAresource units (or resource unit allocation). In some aspects, phaserotation blocks may be provided based on the resource allocationprovided in the second row 910 of FIG. 9 . In one example, the phaserotation blocks may include the following tones, in order from thelowest usable tone index (e.g., −122) to the highest usable tone index(e.g., 122) plus DC tones:

phase rotation block 1 includes 26 data/pilot tones and 1 reserved tone;

phase rotation block 2 includes 26 data/pilot tones and 1 reserved tone;

phase rotation block 3 includes 26 data/pilot tones and 1 reserved tone;

phase rotation block 4 includes 26 data/pilot tones and 1 reserved tone;

phase rotation block 5 includes 13 data/pilot tones, 3 DC tones, and 13data/pilot tones;

phase rotation block 6 includes 1 reserved tone and 26 data/pilot tones;

phase rotation block 7 includes 1 reserved tone and 26 data/pilot tones;

phase rotation block 8 includes 1 reserved tone and 26 data/pilot tones;and

phase rotation block 9 includes 1 reserved tone and 26 data/pilot tones.

This example of the phase rotation blocks may be provided by theshorthand representation as follows, in order from the lowest usabletone index to the highest usable tone index plus DC tones:(26,1)+(26,1)+(26,1)+(26,1)+(13,3,13)+(1,26)+(1,26)+(1,26)+(1,26).

In another example, the phase rotation blocks may include the followingtones, in order from the lowest usable tone index (e.g., −122) to thehighest usable tone index (e.g., 122) plus DC tones:

phase rotation block 1 includes 26 data/pilot tones and 1 reserved tone;

phase rotation block 2 includes 26 data/pilot tones;

phase rotation block 3 includes 26 data/pilot tones and 1 reserved tone;

phase rotation block 4 includes 26 data/pilot tones;

phase rotation block 5 includes 13 data/pilot tones, 7 DC tones, and 13data/pilot tones;

phase rotation block 6 includes 26 data/pilot tones;

phase rotation block 7 includes 1 reserved tone and 26 data/pilot tones;

phase rotation block 8 includes 26 data/pilot tones; and

phase rotation block 9 includes 1 reserved tone and 26 data/pilot tones.

This example of the phase rotation blocks may be provided by theshorthand representation as follows, in order from the lowest usabletone index to the highest usable tone index plus DC tones:(1,26)+(26)+(1,26)+(26)+(13,7,13)+(26)+(26,1)+(26)+(26,1). This may bereferred to as the “Shorthand Representation 20-2.” In this example, theconventions used are as follows: The notation “+” is used to distinguishbetween the different phase rotation blocks. “0” itself may represent aphase rotation block. The sum of the values within each “0” mayrepresent the number of tones for its respective phase rotation block.The summation of all of the number counts equals the total number oftones (e.g., data tone, pilot tone, reserved tone and DC tones).Further, each “26” represents 26 data/pilot tones, each “1” represents 1reserved tone, each “13” represents 13 data/pilot tones, and the “3”represents 3 DC tones. Similar conventions may be used for othershorthand phase rotation block representations provided below, in thateach “52” represents 52 data/pilot tones, each “58” represents 58data/pilot tones, each “59” represents 59 data/pilot tones, each “62”represents 62 data/pilot tones, each “63” represents 63 data/pilottones, each “64” represents 64 data/pilot tones, each “2” represents 2reserved tones, each “14” represents 14 data/pilot tones, and the “5”represents 5 DC tones, and the “7” represents 7 DC tones. In the aboveexample, there are a total of 9 phase rotation blocks, having (a) 8phase rotation blocks that include 27 tones and (b) a central phaserotation block that includes 29 tones (including the 3 DC tones), for atotal of 245 tones.

In some aspects, the phase rotation blocks may be provided based on theresource allocation illustrated in the second row 1010 or the third row1015 of FIG. 10 . In such implementations, the phase rotation block mayinclude the following tones, in order from the lowest usable tone index(e.g., −122) to the highest usable tone index (e.g., 122) plus DC tones:

phase rotation block 1 includes 52 data/pilot tones and 1 reservedtones;

phase rotation block 2 includes 52 data/pilot tones and 1 reservedtones;

phase rotation block 3 includes 13 data/pilot tones, 7 DC tones, and 13data/pilot tones;

phase rotation block 4 includes 1 reserved tones and 52 data/pilottones; and

phase rotation block 5 includes 1 reserved tones and 52 data/pilottones.

This example of the phase rotation blocks may be provided by theshorthand representation of (1,52)+(1,52)+(13,7,13)+(52,1)+(52,1). Inthis example, there are 4 phase rotation blocks that include 54 tonesand a center phase rotation block that includes 29 tones (including the3 DC tones), for a total of 245 tones.

Examples of phase rotation blocks are provided as follows for a 40 MHzchannel bandwidth.

In some aspects, two phase rotation blocks may be utilized (e.g., onephase rotation block for negative tone indices and one phase rotationblock for positive tone indices).

In some aspects, eight phase rotation blocks may be utilized. As oneexample, the phase rotation blocks may include, in order from the lowesttone index (e.g., −244) to the highest tone index (e.g., 244), excludingguard tones: 58 tones, 63 tones, 63 tones, 58 tones, 5 DC tones, 58tones, 63 tones, 63 tones, and 58 tones. A shorthand representation isas follows: (58), (63), (63), (58), (5 DC tones), (58), (63), (63) (58).In this example, the conventions used are as follows: “0” represents aphase rotation block. The value within each “0” represents the number oftones for its respective phase rotation block. Similar conventions maybe used for other shorthand phase rotation block representationsprovided below

As another example, the phase rotation blocks may include, in order, 59tones, 62 tones, 62 tones, 59 tones, 5 DC tones, 59 tones, 62 tones, 62tones, and 59 tones. A shorthand representation is as follows: (59),(62), (62), (59), (5 DC tones), (59), (62), (62) (59).

As another example, the phase rotation blocks may include, in order, 52tones, 64 tones, 64 tones, 64 tones, 64 tones, 64 tones, 64 tones, and53 tones, where DC tones are part of the phase rotation blocks. Ashorthand representation is as follows: (52), (64), (64), (64), (64),(64), (64), (53), where DC tones are part of the phase rotation block.

In some aspects, the phase rotation blocks may be provided based on theresource allocation illustrated in the second row 1110 of FIG. 11 . Insuch implementations, the phase rotation blocks may have 18 phaserotation blocks, and may include the following tones, in order from thelowest tone index (e.g., −244) to the highest tone index (e.g., 244),excluding guard tones:(1,26)+(26,1)+(1,26)+(26,1)+(26)+(1,26)+(26,1)+(1,26)+(26,1), {0, 0, 0,0, 0}, (1,26)+(26,1)+(1,26)+(26,1)+(26)+(1,26)+(26,1)+(1,26)+(26,1). The“{0, 0, 0, 0, 0}” represents the actual DC tone sequence values of 5 DCtones, and they are depicted above simply to show their locations, butthe phase rotation blocks do not include the 5 DC tones. In other words,the 5 DC tones are not phase rotated. Each “26” represents a 26-toneresource unit (e.g., containing 26 data/pilot tones) and each “1”represents a reserved tone. The sum of the values within each “0”represents the number of tones for its respective phase rotation block.For example, (26,1) represents 27 tones (e.g., 26 data/pilot tonesfollowed by 1 reserved tone).

In some aspects, the phase rotation blocks may be provided based on theresource allocation illustrated in the third row 1115 of FIG. 11 . Insuch implementations, the phase rotation block may include the followingtones, in order from the lowest tone index (e.g., −244) to the highesttone index (e.g., 244), excluding guard tones:(1,52,1)+(1,52,1)+(26)+(1,52,1)+(1,52,1), {0, 0, 0, 0, 0},(1,52,1)+(1,52,1)+(26)+(1,52,1)+(1,52,1). The 10 phase rotation blocksdo not include the 5 DC tones, which are depicted within “0” simply toshow their locations. The “52” represents a 52-tone resource unit andthe “1” represents a reserved tone. The sum of the values within each“0” represents the number of tones for its respective phase rotationblock. For example, (1,52,1) represents 54 tones (e.g., 1 reserved tonefollowed by 52 data/pilot tones followed by 1 reserved tone).

Examples of phase rotation blocks are provided as follows for an 80 MHzchannel bandwidth.

In some aspects, five phase rotation blocks may be utilized for theOFDMA case and the non-OFDMA case. For the OFDMA case, the phaserotation blocks may include the following tones, in order from thelowest tone index (e.g., −500) to the highest tone index (e.g., 500),excluding guard tones:

phase rotation block 1 includes 242 data/pilot tones;

phase rotation block 2 includes 242 data/pilot tones;

phase rotation block 3 includes 13 data/pilot tones, 7 DC tones, and 13data/pilot tones;

phase rotation block 4 includes 242 data/pilot tones; and

phase rotation block 5 includes 242 data/pilot tones.

This example of the phase rotation blocks may be provided by theshorthand representation of (242)+(242)+(13,7,13)+(242)+(242). Thisexample uses conventions similar to those describe above. The “242”represents 242 data/pilot tones, the “13” represents 13 data/pilottones, and “7” represents 7 DC tones.

For the non-OFDMA case, the phase rotation blocks may include thefollowing tones, in order from the lowest tone index (e.g., −500) to thehighest tone index (e.g., 500), excluding guard tones:

phase rotation block 1 includes 242 data/pilot tones;

phase rotation block 2 includes 242 data/pilot tones;

phase rotation block 3 includes 14 data/pilot tones, 5 DC tones, and 14data/pilot tones;

phase rotation block 4 includes 242 data/pilot tones; and

phase rotation block 5 includes 242 data/pilot tones.

This example of the phase rotation blocks may be provided by theshorthand representation of (242)+(242)+(14,5,14)+(242)+(242). Thisexample uses conventions similar to those describe above. The “242”represents 242 data/pilot tones, the “14” represents 14 data/pilottones, and “5” represents 5 DC tones.

In some aspects, 16 phase rotation blocks may be utilized for the OFDMAcase and the non-OFDMA case. For the OFDMA case, an example of the phaserotation blocks may be provided by the shorthand representation asfollows, in order from the lowest tone index (e.g., −500) to the highesttone index (e.g., 500), excluding guard tones: (58), (63), (63), (58),(58), (63), (63), (58,13), (13,58), (63), (63), (58), (58), (63), (63),(58), which represents the 994 data/pilot tones but excludes the 7 DCtones. For the non-OFDMA case, an example of the phase rotation blocksmay be provided by the shorthand representation of (58), (63), (63),(58), (58), (63), (63), (58,14), (14,58), (63), (63), (58), (58), (63),(63), (58), which represents the 996 data/pilot tones but excludes the 5DC tones. These example use conventions similar to those describe above.Each number within “0” represents the number of data/pilot tones.

In some aspects, the phase rotation blocks may be provided based on theresource allocation illustrated in the second row 1210 of FIG. 12 . Insuch aspects, 37 phase rotation blocks may be utilized for the OFDMAcase and the non-OFDMA case. For the OFDMA case, an example of the phaserotation blocks may be provided by the shorthand representation asfollows, in order from the lowest tone index (e.g., −500) to the highesttone index (e.g., 500), excluding guard tones:(1,26)+(26,1)+(1,26)+(26,1)+26+(1,26)+(26,1)+(1,26)+(26,1),(1,26)+(26,1)+(1,26)+(26,1)+26+(1,26)+(26,1)+(1,26)+(26,1),(13,{0,0,0,0,0,0,0},13),(1,26)+(26,1)+(1,26)+(26,1)+26+(1,26)+(26,1)+(1,26)+(26,1),(1,26)+(26,1)+(1,26)+(26,1)+26+(1,26)+(26,1)+(1,26)+(26,1).

This example uses conventions similar to those describe above. In thisexample, each “26” represents a 26 tone resource unit, each “1”represents one reserve tone, and “(13,{0,0,0,0,0,0,0},13)” representsthe center resource unit that includes, in order, 13 data/pilot tones, 7DC tones, and 13 data/pilot tones. The “{0,0,0,0,0,0,0}” represents theactual DC tone sequence values of 7 DC tones. In one aspect, a phaserotation block includes the 7 DC tones.

For the non-OFDMA case, an example of the phase rotation blocks may beprovided by the shorthand representation of:(1,26)+(26,1)+(1,26)+(26,1)+26+(1,26)+(26,1)+(1,26)+(26,1),(1,26)+(26,1)+(1,26)+(26,1)+26+(1,26)+(26,1)+(1,26)+(26,1),(14,{0,0,0,0,0},14),(1,26)+(26,1)+(1,26)+(26,1)+26+(1,26)+(26,1)+(1,26)+(26,1),(1,26)+(26,1)+(1,26)+(26,1)+26+(1,26)+(26,1)+(1,26)+(26,1).

This example uses conventions similar to those describe above. In thisexample, each “26” represents a 26 tone resource unit, each “1”represents one reserve tone, and “(14,{0,0,0,0,0},14)” represents thecenter resource unit that includes, in order, 14 data/pilot tones, 5 DCtones, and 14 data/pilot tones. In one aspect, a phase rotation blockincludes the 5 DC tones.

In some aspects, the phase rotation blocks may be provided based on theresource allocation illustrated in the third row 1215 of FIG. 12 . Insuch aspects, 21 phase rotation blocks may be utilized for the OFDMAcase and the non-OFDMA case.

For the OFDMA case, an example of the phase rotation blocks may beprovided by the shorthand representation of(1,52,1)+(1,52,1)+(26)+(1,52,1)+(1,52,1),(1,52,1)+(1,52,1)+(26)+(1,52,1)+(1,52,1)+(13,{0,0,0,0,0,0,0},13)+(1,52,1)+(1,52,1)+(26)+(1,52,1)+(1,52,1),(1,52,1)+(1,52,1)+(26)+(1,52,1)+(1,52,1). This example uses conventionssimilar to those describe above. In this example, each “52” represents a52 tone resource unit, each “2” represents two adjacent reserve tones,and “(13,{0,0,0,0,0,0,0},13)” represents the center resource unit thatincludes, in order, 13 data/pilot tones, 7 DC tones, and 13 data/pilottones. In one aspect, a phase rotation block includes the 7 DC tones.

For the non-OFDMA case, an example of the phase rotation blocks may beprovided by the shorthand representation of(1,52,1)+(1,52,1)+(26)+(1,52,1)+(1,52,1),(1,52,1)+(1,52,1)+(26)+(1,52,1)+(1,52,1)+(14,{0,0,0,0,0},14)+(1,52,1)+(1,52,1)+(26)+(1,52,1)+(1,52,1),(1,52,1)+(1,52,1)+(26)+(1,52,1)+(1,52,1). This example uses conventionssimilar to those described above. In this example, each “52” representsa 52 tone resource unit, each “1” represents a reserve tone, and“(14,{0,0,0,0,0},14)” represents the center resource unit that includes,in order, 14 data/pilot tones, 5 DC tones, and 14 data/pilot tones. Inone aspect, a phase rotation block includes the 5 DC tones.

In one or more implementations, examples of phase rotation blocks for160 MHz channel bandwidth can be obtained through multiple (e.g., eight)duplications of the phase rotation blocks for 20 MHz channel bandwidth,through multiple (e.g., four) duplications of the phase rotation blocksfor 40 MHz channel bandwidth, or through duplications (e.g., two) of thephase rotation blocks for 80 MHz channel bandwidth.

In one or more implementations, a resource allocation for 80+80 MHzchannel bandwidth can be obtained through multiple (e.g., eight)duplications of the phase rotation blocks for 20 MHz channel bandwidth,through multiple (e.g., four) duplications of the phase rotation blocksfor 40 MHz channel bandwidth, or through duplications (e.g., two) of thephase rotation blocks for 80 MHz channel bandwidth.

In one or more implementations, a start tone index and an end tone indexfor a phase rotation block may be provided in different manners. Thesedifferent manners may be due to the varying position and/or number ofreserved indices. In one aspect, since no data/pilot is transmitted atthese reserved indices, a phase rotation block that is applied to thereserved indices does not affect a PPDU transmission.

When there is one reserved tone between two adjacent resource units,such reserved tone may be grouped with its adjacent phase rotation blockto the left of the reserved tone. In another aspect, the reserved tonemay be grouped with its adjacent phase rotation block to the right ofthe reserved tone. In yet another aspect, the reserved tone is notgrouped with any of its adjacent phase rotation blocks.

When there are two reserved tones between two adjacent resource units,such reserved tones may be grouped with their adjacent phase rotationblock to the left of the reserved tones. In another aspect, the reservedtones may be grouped with their adjacent phase rotation block to theright of the reserved tones. In yet another aspect, the left one of thetwo reserved tones may be grouped with its adjacent phase rotation blockto the left of the left one of the two reserved tones; and the right oneof the two reserved tones may be grouped with its adjacent phaserotation block to the right of the right one of the two reserved tones.In yet another aspect, the reserved tones are not grouped with any ofits adjacent phase rotation blocks. Some of these are illustrated inmore detail with respect to FIGS. 13A through 13C and 14A through 14C.

FIGS. 13A through 13C illustrate examples of different options by whichto represent a start tone index and an end tone index for each of thephase rotation blocks. These different options may be utilized with theresource allocation provided in the second row 910 of FIG. 9 forexample. The 9 phase rotation blocks are labeled “Block 1” through“Block 9” in FIGS. 13A through 13C and are applied to data/pilot tones,which are represented as rectangles that span a set of tones. A bracketdenotes a phase rotation block. The start tone index and the end toneindex of the phase rotation blocks are delineated by vertical dottedlines. The reserved tone indices (or null tone indices) are k=−96, −69,−42, −15, 15, 42, 69, and 96, and the DC tone indices are k=−1, 0,and 1. The reserved tones and DC tones are represented as gaps betweenthe data/pilot tones.

FIG. 13A illustrates an example of a first option for representing thestart and end tone indices for each of the phase rotation blocks:

-   -   phase rotation block 1 include tone indices−122≤k≤−96;    -   phase rotation block 2 include tone indices−95≤k≤−70;    -   phase rotation block 3 include tone indices−69≤k≤−43;    -   phase rotation block 4 include tone indices−42≤k≤−17;    -   phase rotation block 5 include tone indices−16≤k≤16;    -   phase rotation block 6 include tone indices 17≤k≤42;    -   phase rotation block 7 include tone indices 43≤k≤69;    -   phase rotation block 8 include tone indices 70≤k≤95; and phase        rotation block 9 include tone indices 96≤k≤122,        where k denotes tone index. For simplicity, a shorthand        representation is provided as follows:        (1,26)+(26)+(1,26)+(26)+(13,7,13)+(26)+(26,1)+(26)+(26,1), which        is the Shorthand Representation 20-2 described previously. This        example uses conventions similar to those described above. Each        “26” represents 26 data/pilot tones, each “1” represents 1        reserved tone, each “13” represents 13 data/pilot tones, and the        “3” represents 3 DC tones. Each “0” represents a phase rotation        block. For the phase rotation blocks 1 through 4, each reserved        tone is grouped with the phrase rotation block to the left of        the reserved tone. For the phase rotation blocks 6 through 9,        each reserved tone is grouped with the phrase rotation block to        the right of the reserved tone.

FIG. 13B illustrates an example of a second option for representing thestart and end tone indices for each of the phase rotation blocks:

phase rotation block 1 include tone indices−122≤k≤−97;

phase rotation block 2 include tone indices−96≤k≤−70;

phase rotation block 3 include tone indices−69≤k≤−43;

phase rotation block 4 include tone indices−42≤k≤−16;

phase rotation block 5 include tone indices−15≤k≤15;

phase rotation block 6 include tone indices 16≤k≤42;

phase rotation block 7 include tone indices 43≤k≤69;

phase rotation block 8 include tone indices 70≤k≤96; and

phase rotation block 9 include tone indices 97≤k≤122.

For simplicity, a shorthand representation is as follows:(26)+(1,26)+(1,26)+(1,26)+(1,13,3,13,1)+(26,1)+(26,1)+(26,1)+(26). Thisexample uses conventions similar to those describe above. No reservedtones are grouped with the phase rotation block 1 and phase rotationblock 9. For phase rotation blocks 2 through 4, each reserved tone isgrouped with the phrase rotation block to the right of the reservedtone. For the phase rotation blocks 6 through 8, each reserved tone isgrouped with the phrase rotation block to the left of the reserved tone.Phase rotation block 5 includes both of its adjacent reserved tones.

FIG. 13C illustrates an example of a third option for representing thestart and end tone indices for each of the phase rotation blocks:

phase rotation block 1 include tone indices−122≤k≤−97;

phase rotation block 2 include tone indices−95≤k≤−70;

phase rotation block 3 include tone indices−68≤k≤−43;

phase rotation block 4 include tone indices−41≤k≤−16;

phase rotation block 5 include tone indices−14≤k≤14;

phase rotation block 6 include tone indices 16≤k≤41;

phase rotation block 7 include tone indices 43≤k≤68;

phase rotation block 8 include tone indices 70≤k≤95; and

phase rotation block 9 include tone indices 97≤k≤122.

For simplicity, a shorthand representation is as follows: (26) 1 (26) 1(26) 1 (26) 1 (13,3,13) 1 (26) 1 (26) 1 (26) 1 (26). Each “26”represents 26 data/pilot tones, each “1” represents 1 reserved tone,each “13” represents 13 data/pilot tones, and the “3” represents 3 DCtones. Each “0” represents a phase rotation block. In this example, noneof the reserved tones is included into any phase rotation block. Thus,the reserved tones are not phase rotated.

FIGS. 14A through 14C illustrate examples of different options by whichto represent a start tone index and an end tone index for each of thephase rotation blocks. These different options may be utilized with theresource allocation provided in the second row 1010 of FIG. 10 forexample. The 9 phase rotation blocks are labeled “Block 1” through“Block 9” in FIGS. 14A through 14C and are applied to data/pilot tones,which are represented as rectangles that span a set of tones. The starttone index and the end tone index are delineated by vertical dottedlines. The reserved tone indices (or null tone indices) are k=−70, −69,−16, −15, 15, 16, 69, and 70, and the DC tone indices are k=−1, 0,and 1. The reserved tones and DC tones are represented as gaps betweenthe data/pilot tones.

FIG. 14A illustrates an example of a first option for representing thestart and end tone indices for each of the phase rotation blocks:

phase rotation block 1 include tone indices−122≤k≤−97;

phase rotation block 2 include tone indices−96≤k≤−70;

phase rotation block 3 include tone indices−69≤k≤−43;

phase rotation block 4 include tone indices−42≤k≤−16;

phase rotation block 5 include tone indices−15≤k≤15;

phase rotation block 6 include tone indices 16≤k≤42;

phase rotation block 7 include tone indices 43≤k≤69;

phase rotation block 8 include tone indices 70≤k≤96; and

phase rotation block 9 include tone indices 97≤k≤122.

For simplicity, a shorthand representation is provided as follows:(26)+(26,1)+(1,26)+(26, 1)+(1,13,3,13,1)+(1,26)+(26,1)+(1,26)+(26). Thisexample uses conventions similar to those described above. This exampledepicts four 2-consecutive reserved tones. The left one of each of the2-consecutive reserved tones is grouped with the phase rotation block tothe left of the left one. The right one of each of the reserved tones isgrouped with the phase rotation block to the right of the right one ofthe reserved tones.

FIG. 14B illustrates an example of a second option for representing thestart and end tone indices for each of the phase rotation blocks:

phase rotation block 1 include tone indices−122≤k≤−97;

phase rotation block 2 include tone indices−96≤k≤−69;

phase rotation block 3 include tone indices−68≤k≤−43;

phase rotation block 4 include tone indices−42≤k≤−15;

phase rotation block 5 include tone indices−14≤k≤14;

phase rotation block 6 include tone indices 15≤k≤42;

phase rotation block 7 include tone indices 43≤k≤68;

phase rotation block 8 include tone indices 69≤k≤96; and

phase rotation block 9 include tone indices 97≤k≤122.

For simplicity, a shorthand representation is provided as follows:(26)+(26,2)+(26)+(26, 2)+(13,3,13)+(2,26)+(26)+(2,26)+(26). This exampleuses conventions similar to those described above. This example depictsfour 2-consecutive reserved tones. Two-consecutive reserved tones aregrouped either with the phase rotation block to the left of the reservedtones or to the right of the reserved tones.

FIG. 14C illustrates an example of a third option for representing thestart and end tone indices for each of the phase rotation blocks:

phase rotation block 1 include tone indices−122≤k≤−97;

phase rotation block 2 include tone indices−96≤k≤−71;

phase rotation block 3 include tone indices−68≤k≤−43;

phase rotation block 4 include tone indices−42≤k≤−17;

phase rotation block 5 include tone indices−14≤k≤14;

phase rotation block 6 include tone indices 17≤k≤42;

phase rotation block 7 include tone indices 43≤k≤68;

phase rotation block 8 include tone indices 71≤k≤96; and

phase rotation block 9 include tone indices 97≤k≤122.

For simplicity, a shorthand representation is provided as follows: (26)(26) 2 (26) (26) 2 (13,3,13) 2 (26) (26) 2 (26) (26). This example usesconventions similar to those described above. Each “26” represents 26data/pilot tones, each “2” represents 2 consecutive reserved tones, each“13” represents 13 data/pilot tones, and the “3” represents 3 DC tones.Each “0” represents a phase rotation block. In this example, none of thereserved tones is included into any phase rotation block. Thus, thereserved tones are not phase rotated.

FIGS. 13A through 13C and FIGS. 14A through 14C illustrate examples bywhich to represent the start and end tone indices of the phase rotationblocks. Other manners by which to represent the start and end toneindices of the phase rotation blocks may also be utilized. In someaspects, due to the use of 1 or 2 reserved tones between adjacentresource units, the tone indices at one or both boundaries may fluctuateby 2.

In one aspect, a phase rotation block includes one or more resourceunits without any reserved tone. In another aspect, a phase rotationblock includes one or more resource units with one reserved tone (e.g.,as either the leftmost tone or the rightmost tone of the phase rotationblock). In another aspect, a phase rotation block includes one or moreresource units with two reserved tones (e.g., as the two leftmost tones,as the two rightmost tones, or as one leftmost tone and one rightmosttone of the phase rotation block).

In one or more implementations, hierarchical phase rotation may beutilized to apply phase rotation. The 40 MHz phase rotation may beprovided such that the phase rotation utilized for 20 MHz, denoted asγ_(20 MHz), is maintained and an additional phase rotation, denoted asγ_(40 MHz), is applied on top of each 20 MHz phase rotation.Accordingly, for 40 MHz, the product of two phase rotation values (e.g.,γ_(20 MHz) γ_(40 MHz)) may be applied for each phase rotation block.Similarly, the 80 MHz phase rotation may be provided such that the phaserotation utilized for 20 MHz and 40 MHz are maintained and an additionalphase rotation, denoted as γ_(80 MHz), is applied on top of each 20 MHzand 40 MHz phase rotation.

FIG. 15 illustrates examples of applying phase rotation hierarchically.A first graph 1505 illustrates applying four phase rotation values totheir respective phase rotation blocks utilized for a 20 MHz channelbandwidth. A bracket denotes a phase rotation block, and a value above abracket denotes the phase rotation value for the phase rotation block. Aphase rotation value is multiplied to each respective element of asequence within the bracket (or the phase rotation block). A secondgraph 1510 illustrates applying two-tier phase rotations to the phaserotation blocks. The first tier applies eight phase rotation values totwo sets of the phase rotation blocks utilized for the 20 MHz channelbandwidth (e.g., created by duplicating the 20 MHz phase rotationblocks). The second tier applies four phase rotation values to fourphase rotation blocks after the eight phase rotations are applied.Alternatively, the two-tier phase rotations can be performed by firstobtaining a product of the two-tier phase rotations and then applyingthe product to the underlying phase rotation blocks (e.g., multiplyingeach of the second tier phase rotation values to a respective one of thefirst tier phase rotation values, and then multiplying each resultingproduct to a respective element of the eight phase rotation blocks).

A third graph 1515 illustrates applying three-tier phase rotations tothe phase rotation blocks. The first tier applies sixteen phase rotationvalues to four sets (e.g., four duplicated sets) of the phase rotationblocks utilized for the 20 MHz channel bandwidth (or two duplicated setsof the phase rotation blocks utilized for the 40 MHz channel bandwidth).The second tier applies eight phase rotation values to the eight phaserotation blocks. The third tier applies four phase rotation values tothe four phase rotation blocks as grouped consecutively. Alternatively,the three-tier phase rotations can be performed by first obtaining aproduct of the three-tier phase rotations and then applying the productto the underlying phase rotation blocks (e.g., multiplying the threetier phase rotation values, the second tier phase rotation values, andthe first tier phase rotation values, and then multiplying the resultingproduct to the four sets of the phase rotation blocks).

For the third graph 1515, the 80 MHz channel bandwidth includes a set oftones 1518 not part of the duplication of the 20 and 40 MHz channelbandwidths. The set of tones 1518 is associated with only the third tierphase rotation, and is not associated with any of the lower tier phaserotation. Thus, a portion of the third tier phase rotation valuesapplicable to the set of tones 1518 is applied to the set of tones 1518.

Although FIG. 15 illustrates an example in which each tier of thehierarchy (e.g., going from the first graph 1505 to the second graph1810, and going from the second graph 1810 to the third graph 1515)utilizes the same phase rotation blocks of [+1, −1, −1, −1], differentphase rotation blocks may be utilized between tiers and within each tierof the hierarchy. Furthermore, fewer or more phase rotation blocks maybe utilized. Furthermore, additional one or more tiers of phase rotationmay be applied.

FIGS. 16A through 16C illustrate examples of applying phase rotationshierarchically. The description from FIG. 15 also applies to FIGS. 16Athrough 16C, with the differences between FIG. 15 and FIGS. 16A through16C described herein. FIG. 16A illustrates the resource allocationprovided in the second row 910 of FIG. 9 with phase rotation blockimplementation provided by phase rotation block option 2 for 20 MHz(e.g.,(26,1)+(26,1)+(26,1)+(26,1)+(13,3,13)+(1,26)+(1,26)+(1,26)+(1,26)).Phase rotation values [+1, −1, −1, −1] may be applied to the first fourphase rotation blocks, and the same phase rotation values [+1, −1, −1,−1] may be applied to the last four phase rotation blocks. The reservedtones (1's) may be allocated to different phase rotation blocks or notallocated to any phase rotation block, as described in more detailabove.

In one or more implementations, consecutive phase rotation blocks thathave the same multiplication value may be consolidated into a singlephase rotation block. For example, the second, third, and fourth phaserotation blocks (each having a phase rotation value of −1) for the 20MHz chancel bandwidth may be formulated and/or implemented as one phaserotation block (having a phase rotation value of −1).

FIG. 16B illustrates an example of a 40 MHz channel bandwidth resourceallocation obtained by creating two duplicated sets of the resourceallocation of FIG. 16A (but without the DC tones of FIG. 16A). See,e.g., the phase rotation block option 3 for the 40 MHz channelbandwidth, which is(26,1)+(26,1)+(26,1)+(26,1)+(26)+(1,26)+(1,26)+(1,26)+(1,26), {0, 0, 0,0, 0}, (26,1)+(26,1)+(26,1)+(26,1)+(26)+(1,26)+(1,26)+(1,26)+(1,26). Thereserved tones (1's) may be allocated to different phase rotation blocksor not allocated to any phase rotation block, as described in moredetail above.

Two-tier phase rotation may be applied. The first tier phase rotationcan use two duplicated sets of the phase rotation values used for FIG.16A. No first tier phase rotation is applied to the resource units 1610and 1615 and the 5 DC tones. The second tier phase rotation can usephase rotation values [+1, −1, −1, −1] applied over four phase rotationblocks. No second tier phase rotation is applied to the 5 DC tones. Asdescribed with respect to FIG. 15 , the first and second tier phaserotations can be multiplied together and then applied to the underlying16 phase rotation blocks.

In some aspects, a portion of resource unit 1610 is associated with thefirst one of the phase rotation blocks at the second tier (e.g., +1phase rotation) and the remaining portion of the resource unit 1610 isassociated with the second one of the phase rotation blocks at thesecond tier (e.g., −1 phase rotation). Similarly, a portion of resourceunit 1615 is associated with a third one of the phase rotation blocks atthe second tier (e.g., +1 phase rotation) and the remaining portion ofthe resource unit 1615 is associated with a fourth one of the phaserotation blocks at the second tier (e.g., −1 phase rotation).

FIG. 16C illustrates an example of an 80 MHz channel bandwidth resourceallocation obtained by creating four sets of the resource allocation ofFIG. 16A (but without the DC tones of FIG. 16A). See, e.g., the phaserotation block option 3 for the OFDMA 80 MHz channel bandwidth, whichis: (26,1)+(26,1)+(26,1)+(26,1)+26+(1,26)+(1,26)+(1,26)+(1,26),(26,1)+(26,1)+(26,1)+(26,1)+26+(1,26)+(1,26)+(1,26)+(1,26),(13,{0,0,0,0,0,0,0},13),(26,1)+(26,1)+(26,1)+(26,1)+26+(1,26)+(1,26)+(1,26)+(1,26),(26,1)+(26,1)+(26,1)+(26,1)+26+(1,26)+(1,26)+(1,26)+(1,26).

The reserved tones (1's) may be allocated to different phase rotationblocks or not allocated to any phase rotation block, as described inmore detail above.

The three-tier phase rotation described with respect to the third graph1505 may be applied in a similar manner. The phase rotation illustratedin FIGS. 16A and 16B may be duplicated for each of the four 242-toneresource units shown in FIG. 16C. The resource allocation of FIG. 16Calso includes an additional 26 data/pilot tones in the center resourceunit 1620. In one aspect, the center resource unit 1620 may beassociated with one or more phase rotation blocks/values introduced atthe third-tier phase rotation (e.g., the second and third phase rotationblocks/values at the third-tier phase rotation), and is not associatedwith any lower tier phase rotation blocks/values. See, e.g., the set oftones 1518 in FIG. 15 .

In one or more implementations, HE LTF sequences of longer lengths maybe obtained based on duplications of shorter length LTF sequences (e.g.,subblocks or sub-sequences). With reference to examples in FIG. 9 , asequence length of 26 and 52 may be designed for the 26-tone resourceunit and the 52-tone resource unit, respectively. Multiple sub-sequences(e.g., 26-length sequences) may form a longer sequence (e.g., a242-length sequence; or a 245-length sequence that includes 3 elementsassociated with DC tones).

In one or more aspects, a PAPR associated with a training sequence(e.g., HE LTF sequence) is generally designed to be lower than a PAPRassociated with random data. In one or more aspects, a PAPR of an HE LTFsequence may be 6.5 dB or lower (e.g., 6 dB, 5.5 dB, 5 dB, 4.5 dB). Inone or more aspects, a PAPR for an HE LTF sequence may be 5 dB or lower.In one or more aspects, each of one or more sub-sequences of an HE LTFsequence is associated with a PAPR that is 6.5 dB or lower. In one ormore aspects, each of one or more sub-sequences of an HE LTF sequence isassociated with a PAPR that is 5 dB or lower. In one aspect, an HE LTFsequence consists of one or more sub-sequences. In one aspect, asub-sequence may correspond to a subblock or a resource unit. In oneaspect, a sub-sequence includes elements associated with a set ofdata/pilot tones or with a set of usable tones. In one aspect, asub-sequence includes successive elements. In one aspect, a sub-sequenceis associated with a set of contiguous tones.

It may be difficult to design long sequences with good PAPR properties.For example, while a 52-length sequence may be constructed byconcatenating two identical 26-length sequences, the 52-length sequenceconstructed in this manner may have PAPR properties that are not as lowas a specifically designed 52-length sequence. However, certain length26 sequences, which have low PAPR by themselves and also have low PAPRwhen utilized to construct a 52-length sequence, may be used to form a52-length sequence by concatenating the two 26-length sequences.

FIGS. 17A through 17D illustrate examples of duplication of a 26-tonesequence for a channel bandwidth 20 MHz resource allocation illustratedin FIG. 10 . In FIGS. 17A through 17D, “A” may represent a 26-length LTFsequence associated with 26 data/pilot tones and “B” may represent a 29length LTF sequence associated with, in order, 13 data/pilot tones, 3 DCtones, and 13 data/pilot tones. Although the LTF sequences utilized inFIGS. 17A through 17D are constructed using duplications of the “A”sequence, other manners by which to construct the LTF sequences may beutilized. In some aspects, a respective phase rotation may be applied toeach “A” sequence and the “B” sequence. In some aspects, instead ofduplicating the “A” sequence, one or more different 26-tone LTFsequences may be utilized.

FIG. 17A illustrates an example of duplication of a 26-tone sequence forthe resource allocation illustrated in the second row 1010 of FIG. 10 .The “A” sequence may be utilized for each of the 26-tone resource units.

FIG. 17B illustrates an example of duplication of a 26-tone sequence forthe resource allocation illustrated in the third row 1015 of FIG. 10 .Two “A” sequences may be duplicated to form a 52-tone sequence thatfills each of the 52-tone resource units.

FIG. 17C illustrates an example of duplication of a 26-tone sequence forthe resource allocation illustrated in the fourth row 1020 of FIG. 10 .In FIG. 17C, 106-tone resource units are utilized. Four “A” sequencesmay be duplicated to form 104 tones of each of the 106-tone resourceunits. The notations c and d indicate additional tones to fill the106-tone resource units. The “c” has two tones, and the “d” has twotones. The c={c₁, c₂} and the d={d₁, d₂} are length two sequences thatare utilized to fill the gaps between two “A” sequences. Each of c₁, c₂,d₁, and d₂ may be +1 or −1, and each may be selected to form a 106-tonesequence associated with lower PAPR.

FIG. 17D illustrates an example of duplication of a 26-tone sequence forthe resource allocation illustrated in the fifth row 1025 of FIG. 10 .In FIG. 17D, a single 242-tone resource unit is utilized. The 242-toneresource unit may include, in order, two “A” sequences, a length twosequence c, two “A” sequences, a length two sequence e, a “B” sequence,a length two sequence f, two “A” sequences, a length two sequence d, andtwo “A” sequences. The c={c₁, c₂}, d={d₁, d₂}, e={e₁, e₂}, and f={f₁,f₂} are length two sequences that are utilized to fill the gapsresulting from the duplication of the A sequence, which is a 26-lengthLTF sequence. Each of c₁, c₂, d₁, d₂, e₁, e₂, f₁, and f₂ may be +1 or−1, and each may be selected to form a 242-tone plus 3 DC-tone sequenceassociated with lower PAPR. In one aspect, each of c₁, c₂, d₁, d₂, e₁,e₂, f₁ and f₂ may be a data/pilot tone, and each may be considered as agap filler.

FIGS. 18A through 18D illustrate examples of duplication of a 26-tonesequence for the channel bandwidth 40 MHz resource allocationillustrated in FIG. 11 . In FIGS. 18A through 18D, “A” may represent a26-length LTF sequence associated with 26 data/pilot tones. In someaspects, a respective phase rotation may be applied to each “A”sequence. In some aspects, instead of duplicating the “A” sequence, oneor more different 26-tone LTF sequences may be utilized. Tone indices−2to +2 represent a 5 DC-tone sequence in each of FIGS. 18A through 18D.

FIG. 18A illustrates an example of duplication of a 26-tone sequence forthe resource allocation illustrated in the second row 1110 of FIG. 11 .The “A” sequence may be utilized for each of the 26-tone resource units.Each of the 26 tone resource units is associated with a respective phaserotation block.

FIG. 18B illustrates an example of duplication of a 26-tone sequence forthe resource allocation illustrated in the third row 1115 of FIG. 11 .Two “A” sequences may be duplicated to form a 52-tone sequence thatfills each of the 52-tone resource units. Each of the 52-tone resourceunits and each of the 26-tone resource units are associated with arespective phase rotation block.

FIG. 18C illustrates an example of duplication of a 26-tone sequence forthe resource allocation illustrated in the fourth row 1120 of FIG. 11 .Four “A” sequences may be duplicated to form each 106-tone resourceunit, with length two sequences c={c₁, c₂}, d={d₁, d₂}, c′={c₁, c₂′},and d′={d₁′, d₂′} utilized to fill the gaps resulting from theduplication of the “A” sequence. Each element in c, d, c′, and d′ may be+1 or −1, and each may be selected to form a 106-tone sequenceassociated with lower PAPR. Each of the 106-tone resource units and eachof the 26-tone resource units are associated with a respective phaserotation block.

FIG. 18D illustrates an example of duplication of a 26-tone sequence forthe resource allocation illustrated in the fifth row 1125 of FIG. 11 .Nine “A” sequences may be duplicated to form each 242-tone resourceunit, with length two sequences c={c₁, c₂}, d={d₁, d₂}, e={e₁, e₂},f={f₁, f₂}, c′={c₁, c₂′}, d′={d₁′, d₂′}, e′={e₁′, e₂′}, and f={f₁′, f₂′}utilized to fill the gaps resulting from the duplication of the “A”sequence. Each of c, d, e, f, c′, d′, e′, and f may be +1 or −1, eachmay be a data/pilot tone, each may be considered a gap filler, and eachmay be selected to form a 242-tone sequence associated with lower PAPR.Each of the 242-tone resource units is associated with a respectivephase rotation block.

FIG. 19A illustrates an example of an LTF sequence for an 80 MHz channelbandwidth constructed using LTF sequences for a 40 MHz channelbandwidth. A resource allocation for the 80 MHz channel bandwidth may bethe resource allocation described for the fifth row 1225 in FIG. 12 .The LTF sequence for 80 MHz channel bandwidth may include four 20 MHzsequences (e.g., 242-tone LTF sequences) and a 33-tone center resourceunit. In some aspects, one or more of the 20 MHz sequences may be usedas shown in FIG. 17D, except with the DC tones (shown in FIG. 17 )removed.

FIG. 19B illustrates an example of an LTF sequence for an 80 MHz channelbandwidth constructed using LTF sequences for a 40 MHz channelbandwidth. A resource allocation for the 80 MHz channel bandwidth may bethe resource allocation described for the seventh row 1235 in FIG. 12 .The LTF for 80 MHz channel bandwidth may include two 40 MHz sequences(e.g., 484-tone LTF sequences), two 14-tone sequences, and one 5-DC tonesequence. The 80 MHz channel bandwidth may be utilized for non-OFDMAcases.

For each of the 20 MHz, 40 MHz, and 80 MHz HE PPDU, different LTFsequences may be provided. The LTF sequence may be selected based onPAPR. Generally, the LTF sequences associated lower PAPR are desirable.Although examples of sequence options are provided below, other sequenceoptions are possible and may be utilized instead.

In some aspects, a portion or an entirety of a VHT LTF sequence may beutilized to construct the HE LTF sequence. For example, a 20 MHz HE PPDUtransmission may utilize the 80 MHz VHT LTF sequence VHTLTF_(−122,122)as the 20 MHz HE PPDU HE LTF sequence, where VHTLTF_(−122,122) isprovided in Equations (7) and (8).

In some aspects, the 20 MHz HE PPDU transmission may utilize thefollowing LTF sequence:HELTF_(2a)={[HELTF₂₆,a₁],[HELTF₂₆,a₂],[HELTF₂₆,a₃],[HELTF₂₆,a₄],[HELTF_(26,1:13),0,0,0,HELTF_(26,14:26)],[a₈,HELTF₂₆],[a₆,HELTF₂₆],[a₇,HELTF₂₆],[a₈,HELTF₂₆]}  Equation(16)where HELTF_(26,1:13) corresponds to the left 13 elements of the HELTF₂₆sequence, HELTF_(26, 14:26) corresponds to the right 13 elements ofHELTF₂₆ sequence. a₁ through as are values for reserved tones, and theyare real values of either +1 or −1. The HELTF₂₆ sequence can beLTF_(left), LTF_(right), or other binary phase shift keying (BPSK)sequence. See Equations (2) and (3) for LTF_(left) and LTF_(right),respectively. The HELTF₂₆ sequence and the a₁ through as may be selectedfor use in the HELTF₂a sequence presented by Equation (16) based on aresulting PAPR of the HELTF₂a sequence. For example, the HELTF₂₆sequence and the a₁ through as may be selected so as to minimize orreduce the resulting PAPR of the HELTF₂a sequence. In some aspects, adifferent sequence for HELTF₂₆ may be utilized for each occurrence ofHELTF₂₆ in Equation (16). This sequence option may be utilized withphase rotation block option 2 for 20 MHz, with the positions of the a₁through as coinciding with the positions of the reserved tones.

In some aspects, the 20 MHz HE PPDU transmission may utilize thefollowing LTF sequence:HELTF_(3a)=[c₁,HELTF₅₂,c₂],[c₃,HELTF₅₂,c₄],[HELTF_(26,1:13),0,0,0,HELTF_(26,14:26)],[c₅,HELTF₅₂,c₆],[c₇,HELTF₅₂,c₈]  Equation(17)where HELTF₅₂ can be {LTF_(left), LTF_(left)}, {LTF_(right),LTF_(right)}, {LTF_(left), LTF_(right)}, or a 52-length BPSK sequence.The c₁ through c₈ are values for reserved tones, and they are realvalues of either +1 or −1. The HELTF₅₂ sequence and the c₁ through c₈may be selected for use in the HELTF₃a sequence presented by Equation(17) based on a resulting PAPR of the HELTF₂a sequence. For example, theHELTF₅₂ sequence and the c₁ through c₈ may be selected so as to minimizeor reduce the resulting PAPR of the HELTF₃a sequence. In some aspects, adifferent sequence for HELTF₅₂ may be utilized for each occurrence ofHELTF₅₂ in Equation (17). This sequence option may be utilized withphase rotation block option 3 for 20 MHz, with the positions of the c₁through c₈ coinciding with the positions of the reserved tones.

In one or more implementations, a base sequence for the 40 MHz HE LTFsequence may be constructed using a portion or an entirety of a VHT LTFsequence. The base sequence may be provided by:HELTF₂₄₂={LTF_(left),1,LTF_(right),−1,−1,−1,1,1,−1,1,−1,1,1,−1,LTF_(left),1,LTF_(right),1,−1,1,−1,1,−1,−1,1,LTF_(left),1,LTF_(right),−1,−1,−1,1,1,−1,1,−1,1,−1,1,1,−1,LTF_(left),1,LTF_(right)}  Equation(18)The HELTF₂₄₂ of Equation (18) includes the 80 MHz VHT LTF sequenceVHTLTF_(−122,122) of Equations (7) and (8) without the center threezeros values, which were mapped to DC tones. The total length of theHELTF₂₄₂ sequence is 242, while the total length of theVHTLTF_(−122,122) sequence is 245.

In some aspects, the 40 MHz HE PPDU transmission may utilize thefollowing sequence: {HELTF₂₄₂, 0, 0, 0, 0, 0, HELTF₂₄₂}. This sequenceoption may be utilized with phase rotation block option 1 for 40 MHz.

In some aspects, the 40 MHz HE PPDU transmission may utilize thefollowing LTF sequence:HELTF_(484A)={[a₁,HELTF₂₆],[HELTF₂₆,a₂],[a₃,HELTF₂₆],[HELTF₂₆,a₄],[HELTF₂₆],[a₈,HELTF₂₆],[HELTF₂₆,a₆],[a₇,HELTF₂₆],[HELTF₂₆,a₈],[0,0,0,0,0],[a₁,HELTF₂₆],[HELTF₂₆,a₂],[a₃,HELTF₂₆],[HELTF₂₆,a₄],[HELTF₂₆],[a₈,HELTF₂₆],[HELTF₂₆,a₆],[a₇,HELTF₂₆],[HELTF₂₆,a₈]}

In some aspects, the 40 MHz HE PPDU transmission may utilize thefollowing LTF sequence:HELTF_(484A)={[a₁,HELTF₂₆],[HELTF₂₆,a₂],[a₃,HELTF₂₆],[HELTF₂₆,a₄],[HELTF₂₆],[a₈,HELTF₂₆],[HELTF₂₆,a₆],[a₇,HELTF₂₆],[HELTF₂₆,a₈],[0,0,0,0,0],[a₉,HELTF₂₆],[HELTF₂₆,a₁₀],[a₁₁,HELTF₂₆],[HELTF₂₆,a₁₂],[HELTF₂₆],[a₁₃,HELTF₂₆],[HELTF₂₆,a₁₄],[a₁₅,HELTF₂₆],[HELTF₂₆,a₁₆]}  Equation(20)In Equations (19) and Equation (20), HELTF₂₆ can be {LTF_(left)},{LTF_(right)}, or a 26-length sequence (e.g., associated with low PAPR).a₁ through a₁₆ are values for reserved tones, and they are real valuesof either +1 or −1. The sequences of Equations (19) and (20) may beutilized with phase rotation block option 3 for 40 MHz, with thepositions of the a₁ through as for Equation (19) and a₁ through a₁₆ forEquation (20) coinciding with the positions of the reserved tones.

In some aspects, the 40 MHz HE PPDU transmission may utilize thefollowing LTF sequence:HELTF_(484B)={[c₁,HELTF₅₂,c₂],[c₃,HELTF₅₂,c₄],[HELTF₂₆],[c₅,HELTF₅₂,c₆],[c₇,HELTF₅₂,c₈],[0,0,0,0,0],[c₁,HELTF₂₆,c₂],[c₃,HELTF₅₂,c₄],[HELTF₂₆],[c₅,HELTF₅₂,c₆],[c₇,HELTF₅₂,c₅]}  Equation(21)

In some aspects, the 40 MHz HE PPDU transmission may utilize thefollowing LTF sequence:HELTF_(484B)={[c₁,HELTF₅₂,c₂],[c₃,HELTF₅₂,c₄],[HELTF₂₆],[c₅,HELTF₅₂,c₆],[c₇,HELTF₅₂,c₈],[0,0,0,0,0],[c₉,HELTF₂₆,c₁₀],[c₁₁,HELTF₅₂,c₁₂],[HELTF₂₆],[c₁₃,HELTF₅₂,c₁₄],[cis,HELTF₅₂,c₁₆]}  Equation(22)In Equations (21) and Equation (22), HELTF₂₆ can be {LTF_(left)},{LTF_(right)}, or a 26-length sequence (e.g., associated with low PAPR).c₁ through c₁₆ are values for reserved tones, and they are real valuesof either +1 or −1. The sequences of Equations (21) and (22) may beutilized with phase rotation block option 4 for 40 MHz, with thepositions of the c₁ through c₈ for Equation (21) and c₁ through c₁₆ forEquation (22) coinciding with the positions of the reserved tones.

For 80 MHz HE PPDU transmission, the LTF sequence utilized for the OFDMAcase and the non-OFDMA case may differ.

In some aspects, for the OFDMA case, the 80 MHz HE PPDU transmission mayutilize the following LTF sequence:HELTF₉₉₄={[HELTF₂₄₂],[HELTF₂₄₂],[HELTF_(26,1:13),0,0,0,0,0,0,0,HELTF_(26,14:26)],[HELTF₂₄₂],[HELTF₂₄₂]}  Equation(23)where HELTF₂₄₂ is provided in Equation (18).

In some aspects, for the non-OFDMA case, the 80 MHz HE PPDU transmissionmay utilize the following LTF sequence:HELTF₉₉₆={[HELTF₂₄₂],[HELTF₂₄₂],[HELTF_(28,1:14),0,0,0,0,0,HELTF_(28,15:28)],[HELTF₂₄₂],[HELTF₂₄₂]}  Equation(24)Each of the sequences of Equations (23) and (24) may be utilized withphase rotation block option 1 for 80 MHz.

In some aspects, the 80 MHz HE PPDU transmission may utilize thefollowing LTF sequence:{[HELTF_(484A)],[HELTF_(26,1:13)],[0,0,0,0,0],[HELTF_(26,14:26)],[HELTF_(484A)]}  Equation(25)where HELTF_(484A) is provided by Equations (19) and (20), but the 5 DCtones in Equations (19) and (20) are not part of HELTF_(484A) whenutilized in Equation (25).

In some aspects, the 80 MHz HE PPDU transmission may utilize thefollowing LTF sequence:[HELTF_(484B)],[HELTF_(26,1:13)],[0,0,0,0,0],[HELTF_(26,14:26)],[HELTF_(484B)]  Equation(26)where HELTF_(484B) is provided by Equations (21) and (22), but the 5 DCtones in Equations (21) and (22) are not part of HELTF_(484B) whenutilized in Equation (26).

In one or more implementations, examples of an LTF sequence for 160 MHzchannel bandwidth can be obtained through multiple (e.g., eight)duplications of an LTF sequence for 20 MHz channel bandwidth, throughmultiple (e.g., four) duplications of an LTF sequence for 40 MHz channelbandwidth, or through duplications (e.g., two) of an LTF sequence for 80MHz channel bandwidth. The DC tone sequences for each of the 20 MHz, 40MHz, and 80 MHz channel bandwidth are removed and not copied into an LTFsequence for 160 MHz channel bandwidth.

In one or more implementations, examples of an LTF sequence for 80+80MHz channel bandwidth can be obtained through multiple (e.g., eight)duplications of an LTF sequence for 20 MHz channel bandwidth, throughmultiple (e.g., four) duplications of an LTF sequence for 40 MHz channelbandwidth, or through duplications (e.g., two) of an LTF sequence for 80MHz channel bandwidth. The DC tone sequences for each of the 20 MHz and40 MHz channel bandwidth are removed and not copied into an LTF sequencefor 80+80 MHz channel bandwidth.

For the 20 MHz, 40 MHz, 80 MHz, 160 MHz, and 80+80 MHz HE PPDUtransmission, in one aspect, LTF sequences are generated based on26-length, 52-length, 106-length, and/or 108-length LTF sequences andreserved tone locations/values.

FIGS. 20 through 31 illustrate examples of PAPR values obtained forvarious LTF sequences utilized for a 20 MHz channel bandwidth. Ahorizontal dimension represents tone indices from −122 to +122.Data/pilot tones are represented as rectangles that span a set of tones.Gaps located in the center of each graph represent DC tones. Other gapsrepresent reserved tones. Each PAPR value represents a PAPR value of itsrespective resource unit (or its respective resource allocation block).A PAPR value is represented in decibel (dB). A base LTF sequence isdepicted within a dotted boundary in its respective figure of FIGS. 20through 31 . The other sequences in a given figure may utilize the samebase sequence, except that the differences may be where/whether reservedtones and gap fillers are used (see, e.g., FIGS. 17A through 18D). Inone or more aspects, a reserved tone may be +1, −1 or j.

In one or more implementations, a 26-length LTF sequence may beconstructed. For example, the 26-length LTF sequence may be utilized orotherwise associated with a 26-tone resource unit.

In one or more implementations, the 26-length LTF sequence may beprovided as follows:LTFah_(−16:15)={0,0,0,1,−1,1,−1,−1,1,−1,1,1,−1,1,1,1,0,−1,−1,−1,1,−1,−1,−1,1,−1,1,1,1,−1,0,0}  Equation(27)from which the following portion of the LTFah_(−16:15) may be obtained:LTFah_(−13:13)={1,−1,1,−1,−1,1,−1,1,1,−1,1,1,1,0,−1,−1,−1,1,−1,−1,−1,1,−1,1,1,1,−1}  Equation(28)The LTFah_(left)=LTFah_(−13:13) (1:13) sequence, referring to the first13 elements of the LTFah_(−13:13) sequence, and theLTFah_(right)=LTFah_(−13:13) (15:27) sequence, referring to the last 13elements of the LTFah_(−13:13) sequence, may be provided.

As a first example, a 26-length LTF sequence may be provided byHELTF₂₆={LTFah_(left), LTFah_(right)}. With reference to the resourceallocation illustrated in the second row 910 of FIG. 9 , the 26-lengthLTF sequence may be followed by or preceded by a reserved tone. In sucha case, a portion of the resource allocation in the second row 910 ofFIG. 9 may include {HELTF₂₆, 1, HELTF₂₆, 1, . . . }, where the “1”represents a reserved tone. With reference to the resource allocationillustrated in the second row 1010 of FIG. 10 , the 26-length LTFsequence may be followed by or preceded by two reserved tones. In such acase, a portion of the resource allocation illustrated in the second row1010 of FIG. 10 may include {HELTF₂₆, HELTF₂₆, 1, 1, HELTF₂₆, HELTF₂₆,1, 1, . . . }.

FIG. 20 illustrates an example of the PAPR for each resource allocation,utilizing the {HELTF₂₆, 1, HELTF₂₆, 1, . . . } sequence. No phaserotation has been applied. FIG. 21 illustrates an example of the PAPRfor each resource allocation, utilizing the {HELTF₂₆, HELTF₂₆, 1, 1,HELTF₂₆, HELTF₂₆, 1, 1, . . . } sequence. No phase rotation has beenapplied.

As a second example, a 27-length LTF sequence may be provided byHELTF₂₇={LTFah_(left), 1, LTFah_(right)}, with the rightmost element ofHELTF₂₇ removed to obtain HELTF₂₆. Other manners by which to obtainHELTF₂₆ may be utilized, such as removing a leftmost element of HELTF₂₇.FIG. 22 illustrates an example of the PAPR for each resource allocation,utilizing the HELTF₂₆ obtained through removing a rightmost element ofHELTF₂₇={LTFah_(left), 1, LTFah_(right)}. No phase rotation has beenapplied.

As a third example, a 26-length LTF sequence may be provided byHELTF₂₆=LTF_(left) or HELTF₂₆=LTF_(right), andHELTF_(26center)={LTFah_(left), 0,0,0, LTFah_(right)} for the centerresource unit. With reference to the resource allocation illustrated inthe second row 910 of FIG. 9 , the 26-length LTF sequence may befollowed by or preceded by a reserved tone. In such a case, a portion ofthe resource allocation in the second row 910 of FIG. 9 may include{HELTF₂₆, 1, HELTF₂₆, 1, . . . }, where the “1” represents a reservedtone. With reference to the resource allocation illustrated in thesecond row 1010 of FIG. 10 , the 26-length LTF sequence may be followedby or preceded by two reserved tone. In such a case, a portion of theresource allocation illustrated in the second row 1010 of FIG. 10 mayinclude {HELTF₂₆, HELTF₂₆, 1, 1, HELTF₂₆, HELTF₂₆, 1, 1, . . . }.

FIG. 23 illustrates an example of the PAPR for each resource allocation,utilizing the {LTF_(left), 1, LTF_(left), 1, . . . } sequence andHELTF_(26center)={LTFah_(left), 0,0,0, LTFah_(right)} for the centerresource unit. No phase rotation has been applied. FIG. 24 illustratesan example of the PAPR for each resource allocation, utilizing the{LTF_(right), 1, LTF_(right), 1, . . . } sequence andHELTF_(26center)={LTFah_(left), 0,0,0, LTFah_(right)} for the centerresource unit. No phase rotation has been applied.

FIG. 25 illustrates an example of the PAPR for each resource allocation,utilizing the {LTF_(left), LTF_(left), 1, 1, LTF_(left), LTF_(left), 1,1, . . . } sequence and HELTF_(26center)={LTFah_(left), 0,0,0,LTFah_(right)} for the center resource unit. No phase rotation has beenapplied. FIG. 26 illustrates an example of the PAPR for each resourceallocation, utilizing the {LTF_(right), LTF_(right), 1, 1, LTF_(right),LTF_(right), 1, 1, . . . } sequence and HELTF_(26center)={LTFah_(left),0,0,0, LTFah_(right)} for the center resource unit. No phase rotationhas been applied.

As a fourth example, a 26-length LTF sequence may be provided byHELTF₂₆=LTF_(left) for odd resource units, HELTF₂₆=LTF_(right) for evenresource units, and HELTF_(26center)={LTFah_(left), 0,0,0,LTFah_(right)} for the center resource unit. The resource units may benumbered using consecutive integers, with the leftmost (e.g., lowestusable tone index) resource unit numbered as resource unit #1. Withreference to the resource allocation illustrated in the second row 910of FIG. 9 , the 26-length LTF sequence may be followed by or preceded bya reserved tone. In such a case, a portion of the resource allocation inthe second row 910 of FIG. 9 may include {LTF_(left), 1, LTF_(right), 1,. . . }, where the “1” represents a reserved tone. With reference to theresource allocation illustrated in the second row 1010 of FIG. 10 , the26-length LTF sequence may be followed by or preceded by two reservedtone. In such a case, a portion of the resource allocation illustratedin the second row 1010 of FIG. 10 may include {LTF_(left), LTF_(right),1, 1, LTF_(left), LTF_(right), 1, 1, . . . }.

In each of the HELTF₂₆ sequences provided above, a “−1” may be used inplace of “1” for each of the reserve tone positions.

In one or more implementations, different LTF sequences may be appliedfor different resource unit location (e.g., tone indices). As apreviously provided example, a 26-length LTF sequence may be provided byHELTF₂₆=LTF_(left) for odd resource units, HELTF₂₆=LTF_(right) for evenresource units, and HELTF_(26center)={LTFah_(left), 0,0,0,LTFah_(right)} for the center resource unit.

A different LTF sequence may be applied for the center resource unit,which may have a different tone mapping due to the DC tones. As anexample, for 26-tone resource units that are not located in the center(e.g., these 26 tone resource units are not the center resource unit),an LTF₂₆ sequence (e.g., HELTF₂₆) may be mapped. For the center resourceunit, an LTF_(center) sequence that has 26 data/pilot tones and 3 zerosat the DC tones may be utilized. For example,LTF_(center)={LTFah_(−13:13) (1:13), 0, 0, 0, LTFah_(−13:13) (15:27)}may be mapped in the center resource unit.

In one or more implementations, a 52-length LTF sequence may be obtainedby extending a length 26 sequence (e.g., by concatenating two identical26-length sequences). Alternatively, a 52-length LTF sequence may beconstructed using existing 20 MHz IEEE 802.11ac LTF sequences. Anexample HELTF₅₂ sequence can be provided byHELTF₅₂={LTF_(left),LTF_(right)}.

For 52 tone resource units, an LTF₅₂ sequence may be mapped, but for thecenter resource unit, an LTF_(center) sequence may be used. An exampleresource allocation may be provided by {1, HELTF₅₂, 1, 1, HELTF₅₂, 1,LTFah_(left), 1, 1, 0, 0, 0, 1, 1, LTFah_(right), 1, HELTF₅₂, 1, 1,HELTF₅₂, 1}, where LTF_(center)={LTFah_(left), 0, 0, 0, 0, 0,LTFah_(right)} and “1” is utilized at the reserved tones.

FIG. 27 illustrates an example of the PAPR for each resource allocation,utilizing the {HELTF₅₂, 1, 1, HELTF₅₂, 1, 1, LTFah_(left), 0, 0, 0,LTFah_(right), 1, 1, HELTF₅₂, 1, 1, HELTF₅₂} sequence. For a 26-toneresource unit (e.g., smaller than 52 tones), 26 tones of the {HELTF₅₂,1, 1, HELTF₅₂, 1, 1, LTFah_(left), 0, 0, 0, LTFah_(right), 1, 1,HELTF₅₂, 1, 1, HELTF₅₂} sequence that correspond to the tone indices ofthe 26-tone resource unit to be filled may be utilized to for the26-tone resource unit. For a 106-tone resource unit, 106 tones of the{HELTF₅₂, 1, 1, HELTF₅₂, 1, 1, LTFah_(left), 0, 0, 0, LTFah_(right), 1,1, HELTF₅₂, 1, 1, HELTF₅₂} sequence that correspond to the tone indicesof the 106-tone resource unit to be filled may be utilized to for the26-tone resource unit. No phase rotation has been applied.

For 106 tone resource units, LTF₁₀₆ may be mapped, but for the centerresource unit, an LTF_(center) for the center resource unit is utilized.

As a first example, a resource allocation may utilizeHELTF₁₀₆={LTF_(left), 1, LTF_(right), LTF_(left), −1, LTF_(right)} and acenter resource allocation LTF_(center)={LTFah_(left), 0, 0, 0,LTFah_(right)}. FIG. 28 illustrates an example of the PAPR for eachresource allocation, utilizing the HELTF₁₀₆={LTF_(left), 1, LTF_(right),LTF_(left), −1, LTF_(right)} and LTF_(center)={LTFah_(left), 0, 0, 0,LTFah_(right)} sequences. No phase rotation has been applied.

As a second example, a resource allocation may utilize HELTF₁₀₆={1,LTF_(left), LTF_(right), LTF_(left), LTF_(right), −1} and a centerresource allocation LTF_(center)={LTFah_(left), 0, 0, 0, LTFah_(right)}.FIG. 29 illustrates an example of the PAPR for each resource allocation,utilizing the HELTF₁₀₆={1, LTF_(left), LTF_(right), LTF_(left),LTF_(right), −1} and LTF_(center)={LTFah_(left), 0, 0, 0, LTFah_(right)}sequences. No phase rotation has been applied.

For 108-tone resource units, LTF₁₀₈ may be mapped, but for the centerresource unit, an LTF_(center) for the center resource unit is utilized.

As a first example, a resource allocation may utilizeHELTF₁₀₈={LTF_(left), 1, 1, LTF_(right), LTF_(left), −1, −1,LTF_(right)} and a center resource allocationLTF_(center)={LTFah_(left), 0, 0, 0, LTFah_(right)}. FIG. 30 illustratesan example of the PAPR for each resource allocation, utilizing theHELTF₁₀₈={LTh_(left), 1, 1, LTF_(right), LTF_(left), −1, −1,LTF_(right)} and LTF_(center)={LTFah_(left), 0, 0, 0, LTFah_(right)}sequences. No phase rotation has been applied.

As a second example, a resource allocation may utilize HELTF₁₀₈={1, 1,LTF_(left), LTF_(right), LTF_(left), LTF_(right), −1, −1} and a centerresource allocation LTF_(center)={LTFah_(left), 0, 0, 0, LTFah_(right)}.FIG. 31 illustrates an example of the PAPR for each resource allocation,utilizing the HELTF₁₀₈={1, 1, LTF_(left), LTF_(right), LTF_(left),LTF_(right), −1, −1} and LTF_(center)={LTFah_(left), 0, 0, 0,LTFah_(right)} sequences. No phase rotation has been applied.

In one or more implementations, a center tone optimization may beperformed on a number of the center tones for an LTF sequence such thatthe PAPR of the LTF sequence is reduced for a center resource unit,e.g., the 26 tone center resource unit illustrated in the second row910, third row 915, and fourth row 920 of FIG. 9 .

For example, the center 11 tones for the VHTLTF_(−122,122) sequenceprovided in Equation (7) are {1, −1, 1, −1, 0, 0, 0, 1, −1, −1, 1}. AHELTF_(−122,122) sequence may be provided by:HELTF_(−122,122)={LTF_(left),1, LTF_(right),−1,−1,−1,1,1,−1,1,−1,1,1,−1,LTF_(left),1, LTF_(right) ,x ₀ ,x ₁ ,x ₂ ,x ₃,0,0,0,x ₄ ,x ₅ ,x ₆ ,x₇,LTF_(left),1,LTF_(right),−1,−1,−1,1,1,−1,1,−1,1,−1,1,1,−1,LTF_(left),1, LTF_(right)}  Equation (29)where the HELTF_(−122,122) sequence is the same as the VHTLTF_(−122,122)sequence except that x₀, x₁, x₂, x₃, x₄, x₅, x₆, and x₇ are to bedetermined. Each of x₀, x₁, x₂, x₃, x₄, x₅, x₆, and x₇ may be +1 or −1,and may be selected such that the resulting HELTF_(−122,122) sequencehas a reduced PAPR.

FIGS. 32 through 34 illustrate examples of center tone optimization.These figures illustrate examples of PAPR values obtained for variousLTF sequences, with different center tone optimization, utilized for a20 MHz channel bandwidth. A horizontal dimension represents tone indicesfrom −122 to +122. Data/pilot tones are represented as rectangles thatspan a set of tones. Gaps located in the center of each graph representDC tones. Other gaps represent reserved tones. Each PAPR valuerepresents a PAPR value of its respective resource unit (or itsrespective resource allocation block). A PAPR value is represented indB.

FIG. 32 illustrates an example of the PAPR associated with theVHTLTF_(−122,122) sequence of Equation (7) without any phase rotationapplied. FIG. 33 illustrates an example of the PAPR associated with theVHTLTF_(−122,122) sequence of Equation (7) with the [1, −1, −1, −1]phase rotation of Equations (13) and (14) applied. The vertical line inFIG. 33 is at tone indexk=−64.

FIG. 34 illustrates an example of the PAPR associated with theHELTF_(−122,122) sequence of Equation (29) with the [1, −1, −1, −1]phase rotation of Equations (13) and (14) applied. The values of x₀, x₁,x₂, x₃, x₄, x₅, x₆, and x₇ are −1, −1, 1, 1, 1, 1, −1, −1, respectively.The vertical line in FIG. 34 is at tone index k=−64. The PAPR values forthe center resource unit in FIG. 34 are lower compared to the PAPRvalues for the center resource units in FIG. 32 and FIG. 33 . Forexample, the PAPR values shown in FIG. 34 for the center resource unitare around 4.4 dB, whereas the PAPR values shown in FIGS. 32 and 33 arearound 7.6 dB.

Although eight values (e.g., x₀, x₁, x₂, x₃, x₄, x₅, x₆, x₇) around theDC tones are selected for optimization in the example above, additional,fewer, or different elements in the VHTLTF_(−122,122) sequence may beselected for optimization. Other values of x₀, x₁, x₂, x₃, x₄, x₅, x₆,and/or x₇ may be utilized.

Since each OFDM symbol duration that utilizes 78.125 kHz tone spacing is12.8 μs long, regular LTF symbols may have a large overhead whentransmitting short frames. In one or more implementations, an LTF symbolduration shorter than 12.8 μs may be utilized. In some aspects, theshorter LTF symbol duration may be implemented by utilizing a smallerbase Fast Fourier Transform (FFT) size. For example, the FFT size may be128 FFT in 20 MHz, compared to 256 FFT in 20 MHz for 78.125 kHz tonespacing. In some aspects, the shorter LTF symbol duration may beimplemented by mapping an LTF sequence in a subset of the tones.

In one or more implementations, 2×LTF transmission may be utilized togenerate two identical 6.4 μs time domain signal replicas (e.g., half of12.8 μs). In 2×LTF transmission, the shorter LTF symbol duration mayutilize 78.125 KHz subcarrier spacing and mapping for every other tone(e.g., even tones or odd tones). The LTF sequence mapping may be on eventones or odd tones. The 2×LTF sequence may be generated by subsampling aregular LTF sequence. A regular LTF sequence may be referred to as a4×LTF sequence. In an aspect, 2×LTF may be referred to as 2×HE LTF. Inan aspect, 4×LTF may be referred to as 4×HE LTF.

In one or more aspects, a 2×LTF sequence may be a subsampled sequence. Asubsampled sequence may include successive elements, where each of thesuccessive elements is associated with a data tone, a pilot tone, or areserved tone. In one aspect, an element of a subsampled sequence is notassociated with a DC tone.

In one or more implementations, a certain number of DC tones may beutilized to reduce issues with channel estimation from local oscillator(LO) leakage. For example, a carrier frequency offset (CFO) of 40parts-per-million (ppm) in 5 GHz may involve at least 7 central DC toneswhile a CFO of 40 ppm in 2.4 GHz may involve at least 3 central DCtones. In some aspects, without a sufficient number of DC tones, areceiver may be configured not to use send data/pilot using the tonesnear the DC, which may have been effected by LO leakage interference.

In one or more implementations, for a 26-tone center resource unit, thesubsampling may leave 12, 13, or 14 tones (e.g., due to presence of DCtones). A 26-length LTF sequence may be utilized as a starting point forconstructing a sequence of 12, 13, or 14 tones. Examples of 26-lengthLTF sequences that may be utilized (e.g., for 26-tone resource units)include LTF_(left), LTF_(right), or LTF_(ah).

FIGS. 35A and 35B illustrate examples of an odd subsampling of a26-length sequence. FIG. 35A illustrates an odd subsampling which may beutilized for a 20 MHz channel bandwidth. FIG. 35B illustrates an oddsubsampling which may be utilized for an 80 MHz OFDMA channel bandwidth.The odd subsampling results in a 2×LTF sequence with 12 non-DC tones.These non-DC tones may be data/pilot tones. In one or moreimplementations, the 12-tone LTF sequence may be generated bysubsampling the odd tones for a 26-length LTF sequence to generate a13-tone LTF sequence, and then truncating one element from the 13-toneLTF sequence to generate the 12-tone LTF sequence. In some aspects, thetruncated element may be a first element or a last element of the13-tone LTF sequence.

FIGS. 36A and 36B illustrate examples of an even subsampling of a26-length sequence. FIG. 36A illustrates an even subsampling which maybe utilized for a 20 MHz channel bandwidth. FIG. 36B illustrates an evensubsampling which may be utilized for an 80 MHz OFDMA channel bandwidth.The odd subsampling results in a 2×LTF sequence with 14 non-DC tones.These non-DC tones may be data/pilot tones. In one or moreimplementations, the 14-tone LTF sequence may be generated bysubsampling the even tones for a 26-length LTF sequence to generate a13-tone LTF sequence, and then adding one element of +1 or −1 to the13-tone LTF sequence to generate the 14-tone LTF sequence. In someaspects, the added element may be added before a first element or aftera last element of the 13-tone LTF sequence.

FIGS. 37 through 40 illustrate examples of LTF tone subsampling fordifferent channel bandwidths. In FIG. 37 , the LTF tone subsampling isperformed such that only even tones at tone indices outside of the DCtones associated with the different channel bandwidths (e.g., 20 MHz, 40MHz, 80 MHz OFDMA, 80 MHz non-OFDMA) are mapped. For 40 MHztransmission, the 2×LTF signal may avoid being affected by LO leakage,which generally affects the tones between k=−3 and k=+3. The totalnumber of LTF tones for the center 26 resource allocation in 20 MHz and80 MHz is identically 14 modulated symbols for 2×LTF. Furthermore, thenumber of non-OFDMA and OFDMA transmission have in total the same numberof LTF modulated symbols.

In FIG. 38 , the LTF tone subsampling is performed such that only oddtones at tone indices outside of the DC tones associated with thedifferent channel bandwidths are mapped (or utilized) for 2×LTF. Thesubsampling of FIG. 38 may allow identical mapping of LTF sequences for2×LTF across all channel bandwidths. The subsampling may also allow moreLTF modulated symbols for the 80 MHz non-OFDMA, which may improvechannel estimation performance for 80 MHz non-OFDMA transmission.

In FIG. 39 , the LTF tone subsampling is performed such that LTFmodulated symbols are adjacent to DC tones for each channel bandwidth.In other words, a 2×LTF sequence is generated in a manner so that thetones immediately adjacent to DC tones are included in the 2×LTFsequence. This subsampling may allow better channel estimation at thecenter of the channel bandwidth, assuming that LO leakage is generallyconfined to the DC tones.

In FIG. 40 , the LTF tone subsampling is performed such that an LTFmodulated symbol is non-adjacent in tone index to the DC tones, exceptfor the 80 MHz OFDMA case. Except for the 80 MHz OFDMA case, a null toneexists between the LTF modulated symbols closest to the DC tones and theDC tones. This may help reduce or avoid an effect of LO leakageinterference on the LTF modulated symbols. For the 80 MHz OFDMA case,the LTF modulated symbols may be mapped right next to the DC tones dueto the higher number of DC tones (e.g., 7) associated with the 80 MHzOFDMA case.

FIG. 41 illustrates a chart for 2×LTF tones. In one or more aspects,subsampling can be provided by reusing a 4×LTF sequence, reusing half ofa 4×LTF, and/or adding or removing one or tones. In some aspects, a2×LTF may be provided by reusing (retrieving) half of the elements of a4×LTF sequence. For example, in FIG. 41 , half of “26” tones of a 4×LTFare reused as “13” tones of a 2×LTF; in other words, in one aspect, a2×LTF having 13 elements (corresponding to 13 tones) are formed bysubsampling or obtaining half of 26 elements of a 4×LTF sequence(corresponding to 26 tones).

In some aspects, a 2×LTF sequence may reuse (or retrieve) a 4×LTFsequence of the same size. For example, in FIG. 41 , “26” tones of a4×LTF are reused as “26” tones of a 2×LTF; in other words, a 2×LTFhaving 26 elements (corresponding to 26 tones) are formed by obtainingor retrieving a 4×LTF sequence having 26 elements (corresponding to 26tones).

In some aspects, one element (corresponding to one tone) may be added orremoved to form a 2×LTF sequence. For example, one element may be addedor removed in the beginning or the end of a 2×LTF sequence as it isbeing formed. In some aspects, such as for a center resource unit (alsoreferred to as a center subblock) or non-OFDMA, a 2×LTF sequence may beformed by adding or subtracting one element (corresponding to one tone)after obtaining half of elements of a 4×LTF sequence (see, e.g., forming“496” or “498” from “994”). In some aspects, a 2×LTF may be formed byadding one or more elements to a 4×LTF sequence (see, e.g., forming “53”or “54” from “52”).

Examples of LTF sequences for 2×LTF are provided below. The subscript 2×below represents a 2×LTF sequence. For example, LTF_(13,2x) is a 2×LTFsequence having 13 elements (or having a length of 13). The subscript 4×below represents a 4×LTF sequence, which is a regular LTF sequence. Forexample, LTF_(26,4x) is a regular LTF sequence having 26 elements (orhaving a length of 26). The notation (a:b) represents tone indices a tob, inclusive. For example, (1:13) represents tone indices 1 to 13,inclusive. LTF_(26,4x)(1:13) represents the first 13 elements of aregular 26-length LTF sequence; in other words, LTF_(26,4x)(1:13)represents a regular HE LTF₂₆ sequence that is on tone indices 1 to 13.

Examples of 2×LTF of length 13 are provided as follows:LTF_(13,2x)=LTF_(26,4x)(1:13) or LTF_(26,4x)(14:26)where half of LTF_(26,4x) can be used for LTF_(13,2x); orLTF_(13,2x)=LTF_(26,4x) (odd indices) or LTF_(26,4x) (even indices),where every other value in LTF_(26,4x) can be used for LTF_(13,2x).

Examples of 2×LTF of length 14 are provided as follows:LTF_(14,2x)={LTF_(13,2x),1} or {1,LTF_(13,2x)} or {LTF_(13,2x),−1} or{−1,LTF_(13,2x)}where a 1 or −1 is inserted at a start or an end of LTF_(13,2x); orLTF_(14,2x)=LTF_(13,2x)(1:6),[1 or −1],LTF_(13,2x)(7:13) orLTF_(13,2x)(1:7),[1 or −1],LTF_(13,2x)(8:13)where a 1 or −1 is inserted in a middle of LTF_(13,2x).

Examples of 2×LTF of length 12 are provided as follows:LTF_(12,2x)=LTF_(13,2x)(1:12) or LTF_(13,2x)(2:13)where a last value or first value of LTF_(13,2x) is removed; orLTF_(12,2x)=LTF_(13,2x)(1:6),LTF_(13,2x)(8:13)where a middle value can be removed from LTF_(13,2x).

Examples of 2×LTF of length 26 are provided as follows:LTF_(26,2x)=LTF₂₆,where the LTF sequence for 4×LTF is reused;LTF_(26,2x)=LTF_(52,4x)(1:26) or LTF_(52,4x)(27:52)where a half of LTF_(52,4x) is used for LTF_(26,2x); orLTF_(26,2x)=LTF_(52,4x) (odd indices) or LTF_(52,4x) (even indices)where every other value in LTF_(52,4x) is used for LTF_(26,2x).

Examples of 2×LTF of length 53 are provided as follows:LTF_(53,2x)={LTF_(52,4x),1} or {1,LTF_(52,4x)} or {LTF_(52,4x),−1} or{−1,LTF_(52,4x)}where a 1 or −1 is inserted at a start or an end of LTF_(52,2x);LTF_(53,2x)=LTF_(106,4x)(1:53) or LTF_(106,4x)(54:106)where a half of LTF_(106,4x) is used for LTF_(53,2x); orLTF_(53,2x)=LTF_(106,4x) (odd indices) or LTF_(106,4x) (even indices)where every other value in LTF_(106,4x) is used for LTF_(53,2x).

Examples of 2×LTF of length 54 are provided as follows:LTF_(54,2x)=LTF_(108,4x)(1:54) or LTF_(108,4x)(55:108)where a half of LTF_(108,4x) is used for LTF_(54,2x);LTF_(54,2x)=LTF_(108,4x) (odd indices) or LTF_(108,4x) (even indices)where every other value in LTF_(108,4x) is used for LTF_(54,2x);LTF_(54,2x)={LTF_(52,4x),[1 or −1],[1 or −1]} or {[1 or −1],[1 or−1],LTF_(52,4x)} or {[1 or −1],LTF_(52,4x),[1 or −1]}where additional 1 or −1 may be added to LTF_(52,4x); orLTF_(54,2x)={LTF_(53,2x),[1 or −1]} or {[1 or −1],LTF_(53,4x)}where additional 1 or −1 may be added to LTF_(53,4x).

An example of 2×LTF of length 121 is provided as follows:LTF_(121,2x)=LTF_(242,4x) (odd indices) or LTF_(242,4x) (even indices)where every other value in LTF_(242,4x) can be used for LTF_(121,2x).

Another example of 2×LTF of length 121 is to reuse a first or secondhalf of LTF_(242,4x). For example, if LTF_(242,4x){VHT_(−122,122, left),VHT_(−122,122), right}, then LTF_(2x,121)=VHT_(−122,122), left orVHT_(122,122), right, whereVHT_(−122,122, left)=VHTLTF_(−122,122)(1;121) andVHT_(−122,122,right)=VHTLTF_(−122,122)(122;245).

Examples of 2×LTF of length 120 are provided as follows:LTF_(120,2x)=LTF_(121,2x)(1:120) or LTF_(121,2x)(2:121)where a first or last value of LTF_(121,2x) is removed; orLTF_(120,2x)={LTF_(121,2x)(1:60),LTF_(121,2x)(62:121)}where a middle value of LTF_(121,2x) is removed.

Examples of 2×LTF of length 122 are provided as follows:LTF_(122,2x)={LTF_(121,2x),1} or {1,LTF_(121,2x)} or {LTF_(121,2x),−1}or {−1,LTF_(121,2x)}where a 1 or −1 is added at a start or an end of LTF_(121,2x);LTF_(122,2x)={LTF_(121,2x)(1:60),[1 or −1],LTF_(121,2x)(61:121)} or{LTF_(121,2x)(1:61),[1 or −1],LTF_(121,2x)(62:121)}where a 1 or −1 is added in a middle of LTF_(121,2x); orLTF_(122,2x)={LTF_(120,2x),[1 or −1],[1 or −1]} or {[1 or −1],[1 or−1],LTF_(120,2x)} or {[1 or −1],LTF_(120,2x),[1 or −1]}where additional 1 or −1 may be added to LTF_(120,2x).

Examples of 2×LTF of length 242 are provided as follows:LTF_(242,2x)=LTF_(242,4x)where the LTF sequence for 4×LTF can be reused;LTF_(242,2x)=LTF_(484,4x)(1:242) or LTF_(484,4x)(243:484)where a half of LTF_(484,4x) is used for LTF_(242,2x); orLTF_(242,2x)=LTF_(484,4x) (odd indices) or LTF_(484,4x) (even indices)where every other value in LTF_(484,4x) can be used for LTF_(242,2x).

Examples of 2×LTF of length 241 are provided as follows:LTF_(241,2x)=LTF_(242,2x)(1:241) or LTF_(242,2x)(2:242)where a first or last value of LTF_(242,2x) is removed; orLTF_(241,2x)={LTF_(242,2x)(1:120),LTF_(242,2x)(122:242)} or{LTF_(242,2x)(1:121),LTF_(242,2x)(123:242)}where a middle value of LTF_(242,2x) is removed.

Examples of 2×LTF of length 243 are provided as follows:LTF_(243,2x)={LTF_(242,2x),1} or {1,LTF_(242,2x)} or {LTF_(242,2x),−1}or {−1,LTF_(242,2x)}where a 1 or −1 is inserted at a start or an end of LTF_(242,2x); orLTF_(243,2x)={LTF_(242,2x)(1:121),[1 or −1],LTF_(242,2x)(122:242)}where a 1 or −1 is inserted in a middle of LTF_(242,2x).

Examples of 2×LTF of length 497 are provided as follows:LTF_(497,2x)=LTF_(994,4x)(1:497) or LTF_(994,4x)(498:994)where half of LTF_(994,4x) is used for LTF_(497,2x);LTF_(497,2x)=LTF_(994,4x) (odd indices) or LTF_(994,4x) (even indices)where every other value of LTF_(994,4x) is used for LTF_(497,2x);LTF_(497,2x)=LTF_(498,2x)(1:497) or LTF_(498,2x)(2:498)where a first or last value of LTF_(498,2x) is removed; orLTF_(497,2x)={LTF_(498,2x)(1:248),LTF_(498,2x)(250:498)} or{LTF_(498,2x)(1:249),LTF_(498,2x)(251:498)}where a middle value of LTF_(498,2x) is removed.

Examples of 2×LTF of length 496 are provided as follows:LTF_(496,2x)LTF_(497,2x)(1:496) or LTF_(497,2x)(2:497)where a first or last value of LTF_(497,2x) is removed; orLTF_(496,2x)={LTF_(497,2x)(1:248),LTF_(497,2x)(250:497)}where a middle value of LTF_(497,2x) is removed.

Examples of 2×LTF of length 498 are provided as follows:LTF_(498,2x)={LTF_(497,2x),[1 or −1]} or {[1 or −1],LTF_(497,2x)}where a 1 or −1 is inserted at a start or an end of LTF_(497,2x);LTF_(498,2x)={LTF_(497,2x)(1:248),[1 or −1],LTF_(497,2x)(249:497)} or{LTF_(497,2x)(1:249),(1 or −1),LTF_(497,2x)(250:497)}where a 1 or −1 is inserted in a middle of LTF_(497,2x);LTF_(498,2x)=LTF_(996,4x)(1:498) or LTF_(996,4x)(499:996)where half of LTF_(996,4x) is used for LTF_(498,2x); orLTF_(498,2x)=LTF_(996,4x) (odd indices) or LTF_(996,4x) (even indices)where every other value in LTF_(996,4x) is used for LTF_(498,2x).

Examples of 2×LTF of length 499 are provided as follows:LTF_(499,2x)={LTF_(498,2x),[1 or −1]} or {[1 or −1],LTF_(498,2x)}where a 1 or −1 is inserted at a start or an end of LTF_(498,2x); orLTF_(499,2x){LTF_(498,2x)(1:249),[1 or −1],LTF_(498,2x)(250:498)}where a 1 or −1 is inserted in a middle of LTF_(498,2x).

In the examples provided above, in one aspect, LTF_(26,4x) is HELTF₂₆(or LTF₂₆). In one aspect, HELTF₂₆ (or LTF₂₆) can be LTh_(left),LTF_(right), or other binary phase shift keying (BPSK) sequence. In oneaspect, LTF_(52,4x) is HELTF₅₂. In one aspect, LTF_(106,4x) is HELTF₁₀₆.In one aspect, LTF_(108,4x) is HELTF₁₀₈. In one aspect, LTF_(242,4x) isHELTF₂₄₂. In one aspect, LTF_(484,4x) is HELTF_(484A) or HELTF_(484B).In one aspect, LTF_(994,4x) is HELTF₉₉₄. In one aspect, LTF_(996,4x) isHELTF₉₉₆.

In one or more implementations, an HT LTF sequence may be provided fordownlink-OFDMA (DL-OFDMA) and uplink-OFDMA (UL-OFDMA). In DL-OFDMA, aSTA (e.g., an AP 111 of FIG. 1 ) may transmit the HE LTF sequences forall corresponding STAs (e.g., 112-115 of FIG. 1 ) and the AP maytransmit one or more resources to the STAs. In UL-OFDMA, a STA maytransmit an HE LTF sequence (or a portion thereof) that corresponds onlyto those resources for which the STA has been assigned. The transmittedsignals from multiple STAs may make up the whole LTF sequence (orsymbols) at the receiver (e.g., AP 111). In one or more implementations,a STA that is allocated to a center resource unit may not be allocatedto any of the other resource units (e.g., STA4 in FIG. 42 ).

In one or more implementations, an HE LTF sequence length that isutilized is based on the operating bandwidth of OFDMA. The operatingbandwidth may be the transmission bandwidth of the legacy preambleportion or the bandwidth that encompasses all the allocated resources ofOFDMA. In one aspect, an operating bandwidth may be a channel bandwidth.

FIG. 42 illustrates an example of an HE LTF transmission for UL-OFDMAand DL-OFDMA. In UL-OFDMA, each STA (e.g., STA1 through STA5) transmitsonly a subset of the LTF sequence defined over the entire operatingbandwidth. As shown in FIG. 42 , STAs 1 through 5 are allocated with 48,24, 24, 24, and 102 data subcarriers (or data tones), respectively, thatare located within tone indices between −122 and −71, between −68 and−43, between −42 and −17, between −14 to −2 & +2 to +14 (without 3 DCtones at −1, 0, +1), and between +17 and +122, respectively. Each STAtransmits an LTF sequence for only those tones allocated to the STA.

In one aspect, in DL-OFDMA, the AP transmits an HE LTF sequence onlyover the allocated resources (or the allocated tones). For example,although an HE LTF sequence for 20 MHz may be provided for 242 usabletones, the AP transmits an HE LTF sequence associated with only thosetones allocated for the STAs (e.g., STAs that participate in thecommunication) and omits the remaining tones (e.g., reserved tones) fromtransmission. In this example, the AP (e.g., IFT 284 of the AP)uses/modulates symbols associated with the data/pilot tones allocated tothe STAs for transmission but does not use the DC tones or the reservedtones for transmission.

FIG. 43 illustrates an example of an LTF transmission for UL-OFDMA andDL-OFDMA. The UL-OFDMA operates similarly to that illustrated in FIG. 42. In DL-OFDMA, the AP transmits an HE LTF sequence over the entireoccupied data/pilot tones and reserved tones. In this example, the AP(e.g., IFT 284 of the AP) uses/modulates symbols (elements of asequence) associated with the data/pilot tones allocated to the stationsas well as the reserved tones for transmission, but does not use the DCtones for transmission.

FIG. 44 illustrates an example of an LTF transmission for UL-OFDMA andDL-OFDMA. The LTF sequence length that is utilized by a STA is based onthe allocated resource size of OFDMA. An LTF sequence may be providedfor each resource allocation size, such as 26, 52, 106 (or 108), and 242for 20 MHz operations.

In a UL-OFDMA case, each STA transmits only the LTF sequence providedfor its allocated resource size. In some aspects, the number of tones(or an LTF sequence length) allocated to one STA may be different fromthat of another STA. In some aspects, the number of tones allocated totwo or more STAs may be the same. In one aspect, tones allocated to aSTA may be referred to as a subband allocated to the STA. In FIG. 44 ,STA1 is allocated with an LTF sequence having 52 elements (or having alength of 52), which may be referred to as a 52-length LTF sequence. Inone aspect, the elements are associated with tones or data/pilot tones.STA2 is allocated with a 26-length LTF sequence. STA3 is allocated witha 26-length LTF sequence. STA4 is allocated with a 26-length LTFsequence for a center resource unit, where the 3 DC tones are not usedfor transmission. STA5 is allocated with a 106-length LTF sequence or a108-length LTF sequence. A STA (e.g., IFT 284 of the STA) mayuse/modulate symbols (elements of a sequence) associated with tonesallocated to the STA (e.g., data/pilot tones allocated to the station aswell as any reserved tones allocated to the station).

In a DL-OFDMA case, the AP concatenates, into one downlink LTF sequence,the LTF sequences associated with all of the STAs that participate inthe communication. In this example, the AP concatenates the LTFsequences of STA1, STA2, STA3, STA4, and STA5. In other words, in oneaspect, the AP concatenates the 52-length LTF sequence, the 26-lengthLTF sequence, the 26-length LTF sequence, the 26-length LTF sequence forthe center resource unit, and the 106/108-length LTF sequence. In oneaspect, an AP aggregates the LTF sequences allocated to the stationsinto a downlink LTF sequence.

In some aspects, in the DL-OFDMA case, phase rotation may be multipliedper allocated contiguous resource. For example, a phase rotation valuefor each of the 52-tone subblock, the 26-tone subblock, the 26-tonesubblock, the 26-tone center subblock, and the 106/108-tone subblock maybe multiplied to a respective element of the sequences. The phaserotation may allow for a reduction in the transmission PAPR. In thisexample, each LTF sequence assigned to a STA may be considered to be aphase rotation block, and phase rotation may be applied to each suchblock. In addition, in this example each LTF sequence assigned to a STAis associated with a resource unit.

In one or more implementations, the DL-OFDMA and UL-OFDMA may utilizedifferent LTF sequence transmission methods. For example, the DL-OFDMAmay utilize the DL-OFDMA LTF transmission illustrated in FIG. 43 whereasthe UL-OFDMA may utilize the UL-OFDMA LTF transmission illustrated inFIG. 44 .

In one or more implementations, in DL, the AP may transmit the whole HELTF sequence in the entire operating bandwidth. For example, even in acase where only one STA is present and is allocated to a single subband,where the single subband is smaller than the entire operating bandwidth,the AP may transmit the HE LTF sequence for the entire operatingbandwidth. In UL, in one aspect, each STA may transmit only its LTFsequence corresponding to the allocated subband. For a STA, utilizationof different LTF sequences depending on the allocated subband size mayallow for a reduction of PAPR. For example, a STA may get a ULallocation in a 26-tone subband, in which case it may use a 26-lengthLTF sequence separately designed for length 26. When a STA gets a ULallocation in a 52-tone subband and/or a 106-tone (or 108-tone) subband,the STA may use a 52-length LTF sequence and/or a 106-length (or108-length) LTF sequence separately designed for the respective length.

A subband is a portion of a channel bandwidth. For example, a 40 MHzchannel bandwidth may have eighteen 26-tone subbands. An 80 MHz channelbandwidth may have four 242-tone subbands and one 26-tone subband.

In one aspect, a subband may correspond to a resource unit. In oneaspect, the size of a subband may be the size of a resource unit size,or the bandwidth of a resource unit (e.g., 26-tones, 52-tones,106-tones, 108-tones, 242-tones, 484-tones, 994-tones, or 996-tones). Inanother aspect, a subband may have a size, or a bandwidth, differentfrom the size of a resource unit.

In one or more aspects, each STA may be allocated with (or associatedwith) a respective one of the subbands, and the STA's HE LTF sequence isassociated with the respective one of the subbands. For example, when anallocated subband for a STA is a 52-tone subband, the STA's HE LTFsequence is a 52-length sequence (associated with 52 tones). Such52-length sequence may be specifically designed to have a low PAPR forthat length.

Examples of HE LTF sequences for differently-sized subbands are providedbelow. In one aspect, each of the examples provided below may be used asan HE LTF sequence for a UL frame. In another aspect, one or moreexamples provided below may be used as one or more subblocks of an HELTE sequence for a UL frame. In another aspect, one or more examplesprovided below may be used as one or more subblocks of an HE LTEsequence for a DL frame. In one or more aspects, a subblock maycorrespond to a resource unit and vice versa.

An example of a 26-tone sequence for a center subblock can beHELTF_(26center)={LTFah_(left), 0, 0, 0, LTFah_(right)}. Note that26-tone may be referred to as 26-length. The number of DC tones may varydepending on the channel bandwidth. For example, for 40 MHz,HELTF_(26center)={LTFah_(left), 0, 0, 0, 0, 0, LTFah_(right)}, whereasfor 80 MHz OFDMA, HELTF_(26center)={LTFah_(left), 0, 0, 0, 0, 0, 0, 0,LTFah_(right)}.

Examples of a 26-tone sequence for other subblocks (e.g., non-centersubblocks) can be:HELTF₂₆={LTFah_(left),LTFah_(right)};HELTF₂₆={LTF_(left)}; orHELTF₂₆={LTF_(right)}.In some cases, HELTF₂₆ may be changed depending on the tone indices of asubblock. For example, even subblocks may utilize LTF_(left) whereas oddsubblocks may utilize LTF_(right). In some aspects, orthogonality may bemaintained with different size of subblocks in overlapping BSS (OBSS).

In the above examples, the length 13 sequences that are utilized are theLTFah_(left) and LTFah_(right). Other length 13 sequences may beutilized.

An example of a 52-tone sequence for a subblock can beHELTF₅₂={LTF_(left), LTF_(right)}.

Examples of a 106-tone LTF sequence for a subblock can be:HELTF₁₀₆={LTF_(left),1,LTF_(right),LTF_(left),−1,LTF_(right)};HELTF₁₀₆={1,LTF_(left),LTF_(right),LTF_(left),LTF_(right),−1}; orHELTF₁₀₆={LTF_(left),LTF_(right),1,−1,LTF_(left),LTF_(right)},where 1 and −1 are provided as examples.

Examples of a 108-tone LTF sequence for a subblock can be:HELTF₁₀₈={LTF_(left),1,1,LTF_(right),LTF_(left),−1,−1,LTF_(right)};HELTF₁₀₈={1,1,LTF_(left),LTF_(right),LTF_(left),LTF_(right),−1,−1};HELTF₁₀₈={LTF_(left),LTF_(right),1,1,LTF_(left),−LTF_(right),1,−1}; orwhere 1 and −1 are provided as examples.

Examples of a 242-tone LTF sequence for a subblock can be:HELTF₂₄₂={VHT_(−122,122,left),VHT_(−122,122,right)};HELTF₂₄₂={1,HELTF₁₀₆,1,HELTF₂₆,−1,HELTF₁₀₆,−1}; orHELTF₂₄₂={HELTF₁₀₈,HELTF₂₆,HELTF₁₀₈}.A phase rotation may be applied for every one fourth of the HELTF₂₄₂sequence. For example, a phase rotation of [1, −1, −1, −1] may beapplied.

An example of a 484-tone LTF sequence for a subblock can be:HELTF₄₈₄={HELTF₂₄₂, HELTF₂₄₂}. A phase rotation may be applied for theHELTF₄₈₄ sequence.

In one or more implementations, a 994-tone LTF sequence may beconstructed by combinations of HELTF₂₄₂ and HELTF₂₆. For the 26-tonesubblock, which is located in the center of 80 MHz,LTF_(26center)={LTF_(26left), 0, 0, 0, 0, 0, 0, 0, LTF_(26right)} may beutilized as the LTF sequence, where LTF_(26left) is the leftmost length13 sequence in the HELTF₂₆ and LTF_(26right) is the rightmost length 13sequence in the HELTF₂₆.

A 994-tone LTF sequence, HELTF₉₉₄, may be generated using four HELTF₂₄₂sequences and one LTF_(26center) sequence. For example,HELTF₉₉₄={HELTF₂₄₂, HELTF₂₄₂, LTF_(26center), HELTF₂₄₂, HELTF₂₄₂}. Aphase rotation may be applied to each of the HELTF₂₄₂ sequences and theLTF_(26center) sequence. Applying a phase rotation of [1, −1, 1, −1, −1]results in LTF₉₉₄={HELTF₂₄₂, −HELTF₂₄₂, LTF_(26center), −HELTF₂₄₂,−HELTF₂₄₂}.

In one or more implementations, a 996-tone LTF sequence may beconstructed by combinations of HELTF₂₄₂ and HELTF₂₆. For the 26-tonesubblock, which is located in the center of 80 MHz,LTF_(26center)={LTF_(26left), 0, 0, 0, 0, 0, LTF_(26right)} may beutilized as the LTF sequence.

A 996-tone LTF sequence, HELTF₉₉₆, may be generated using four HELTF₂₄₂sequences, one LTF_(26center) sequence, and additional two 1 or −1.Examples of 996-tone LTF sequence may be provided as follows:HELTF₉₉₆={HELTF₂₄₂,[1 or −1],HELTF₂₄₂,LTF_(26center),HELTF₂₄₂,[1 or−1],HELTF₂₄₂};HELTF₉₉₆={[1 or−1],HELTF₂₄₂,HELTF₂₄₂,LTF_(26center),HELTF₂₄₂,HELTF₂₄₂,[1 or −1]}; orHELTF₉₉₆={HELTF₂₄₂,HELTF₂₄₂,[1 or −1],LTF_(26center),[1 or−1],HELTF₂₄₂,HELTF₂₄₂}For each of the example LTF₉₉₆ sequences, phase rotation may be applied.

In one or more aspects, an HE LTF sequence may be formed by duplicatingone or more short sequences (e.g., duplicating one or more shortsequences of resource units). For example, a 994 or 996-length LTFsequence may be formed by duplicating one or more shorter LTF sequences(e.g., 484-length sequence, 242-length sequence, 106/108-lengthsequence, 52-length sequence, and/or 26-length sequence). A 484-lengthLTF sequence may be formed by duplicating one or more shorter LTFsequences (e.g., 242-length sequence, 106/108-length sequence, 52-lengthsequence, and/or 26-length sequence). A 242-length LTF sequence may beformed by duplicating one or more shorter LTF sequences (e.g.,106/108-length sequence, 52-length sequence, and/or 26-length sequence).A 106 or 108-length LTF sequence may be formed by duplicating one ormore shorter LTF sequences (e.g., 52-length sequence and/or 26-lengthsequence). A 52-length LTF sequence may be formed by duplicating one ormore shorter LTF sequences (e.g., 26-length sequence).

In one aspect, a length may refer to the number of tones in a sequence,and vice versa. For example, a 52-length LTF sequence may refer to a52-tone LTF sequence, an LTF sequence with 52 tones or vice versa. Inone aspect, a length may refer to the number of data tones and pilottones in a sequence. In another aspect, a length may refer to the numberof data tones, pilot tones and reserved tones in a sequence. In anotheraspect, a length may refer to the number of data tones, pilot tones,reserved tones and DC tones in a sequence. In one aspect, a length mayrefer to the number of usable tones in a sequence. In another aspect, alength may refer to the number of usable tones and DC tones in asequence. In one aspect, a length may refer to the number elements in asequence.

In one or more aspects, a sequence (or an HE LTF sequence) includes orconsists of elements. In one aspect, each element of a sequence isassociated with a data tone or a pilot tone. In another aspect, eachelement of a sequence is associated with a data tone, a pilot tone, or areserved tone. In another aspect, each element of a sequence isassociated with a data tone, a pilot tone, a reserved tone, or a DCtone. In one aspect, each element of a sequence is associated with atone. In one aspect, elements of a sequence are not associated with anyguard tone. In one aspect, each element of a sequence is associated witha usable tone. In one aspect, each element of a sequence is associatedwith a usable tone or a DC tone. In one aspect, elements are sequential.In one aspect, elements are successive. In one aspect, elements areconsecutive. In one aspect, an element of a sequence is referred to as asymbol, a modulated symbol, an LTF modulated symbol, or vice versa. Inone aspect, a symbol is associated with a tone.

When a long sequence is formed by duplicating one or more shortsequences, and when such short sequences contain DC tones, such DC tonesin the short sequences are removed when forming the long sequence.Further, when a long sequence is formed by duplicating one or more shortsequences, applicable DC tones (e.g., 3, 5 or 7 DC tones) may be addedto the center portion of the long sequence. Further, when a longsequence is formed by duplicating one or more short sequences,applicable reserved tones may be added to form the long sequence.

In one or more aspects, an HE LTF sequence may be identified using asequence including its associated DC tones or may be identified using asequence without its associated DC tones. In one aspect, the notationHELTF_(x) may refer to an HE LTF sequence having x number of usabletones, where x is a number such as 26, 52, 106, 108, 242, 484, 994, or996. In another aspect, the notation HELTF_(x) may refer to an HE LTFsequence having x number of usable tones plus its DC tones. In oneaspect, LTF may refer to HELTF_(x). In one aspect, HELTF₄₈₄ may refer toa sequence of 484 usable tones or to a sequence of 484 usable tones plusits DC tones. In one aspect, HELTF_(484A) may refer to an HE LTFsequence having 484 usable tones or to an HE LTF sequence having 484usable tones plus its DC tones. In one aspect, HELTF_(484B) may refer toan HE LTF sequence having 484 usable tones or to an HE LTF sequencehaving 484 usable tones plus its DC tones.

In one or more aspects, a sequence may refer to one or more sequences.In one or more aspects, a sequence may refer to a portion of a sequence,a subblock of a sequence, one or more subblocks of a sequence, asub-sequence, one or more sub-sequences, or vice versa. In one aspect, asubblock is associated with a resource unit. In one aspect, when asequence includes only one sub-sequence, the sequence and thesub-sequence may be the same.

In one or more aspects, a downlink is directed from at least one STA(e.g., an AP 111) to a plurality of STAs (e.g., 112-115), and an uplinkis directed from the plurality of STAs (e.g., 112-115) to at least oneSTA (e.g., AP 111). In one or more implementations, a processor (e.g.,210, 215, or 280) of a STA (e.g., an AP or a non-AP STA) may determine achannel bandwidth. A processor may also generate, determine, and/orprovide an HE LTF sequence. A processor may also generate or provide aframe (e.g., a downlink frame or a uplink frame) and facilitatetransmission of the frame.

The processor may generate an HE LTF sequence based on certaininformation or may determine and/or retrieve the HE LTF sequence fromstorage (e.g., memory or register(s)) for usage. The processor mayperform phase rotation. The HE LTF sequence may include a centersubblock (or a center sub-sequence) that has been optimized for PAPR. Inone aspect, such center sequence may be different from another subblockof the subsequence. The HE LTF sequence may include two or moresubblocks that are the same.

The processor may subsample a first HE LTF sequence to generate asubsampled HE LTF sequence, where the subsampled HE LTF sequence hasless number of elements than the number of elements of the first HE LTFsequence. In one aspect, subsampling may include reusing/retrieving oneor more subblocks or reusing/retrieving one half of each of one or moresubblocks of the first HE LTF sequence. In one aspect, sub sampling mayinclude adding or removing one tone (e.g., at the beginning or at theend of the subsampled subblock).

In one aspect, to prepare for transmission, the processor may selectonly the portion(s) of the sequence associated with the data/pilot tonesthat are allocated to the STAs participating in the communication. Inanother aspect, the processor may select portion(s) of the sequenceassociated with the data/pilot tones allocated to the STAs and reservedtones. In yet another aspect, the processor may concatenate a set ofsequences allocated to the STAs.

In one or more implementations, a processor (e.g., 210, 215, or 280) ofa STA (e.g., a non-AP STA) may generate, determine or provide an HE LTFsequence having a length that is less than a channel bandwidth. In oneaspect, the HE LTF sequence has a length of a resource unit.

FIG. 45 shows examples of structures of HE LTF sequences in 1× HE LTFmode for 20 MHz, 40 MHz, and 80 MHz channel bandwidths. The HE-LTFsequences shown in FIG. 45 can be used for both non-OFDMA transmissionand OFDMA transmission. In an aspect, 1×HE LTF may be referred to as1×LTF.

The HE-LTF sequence for the 20 MHz channel bandwidth and the 1×HE-LTFmode may include zero values on subcarrier indices from −122 to −4excluding every fourth subcarrier index from −122 to −4, non-zero valueson every fourth subcarrier index from −122 to −4, zero values onsubcarrier indices from −3 to −2, DC tones on subcarrier indices from −1to 1, zero values on subcarrier indices from 2 to 3, non-zero values onevery fourth subcarrier index from 4 to 122, and zero values onsubcarrier indices from 4 to 122 excluding every fourth subcarrier indexfrom 4 to 122.

For the HE-LTF sequence for the 20 MHz and the 1×HE-LTF mode, annon-limiting example of the sequence of non-zero values is as follows:{x ₁ ,x ₂,1,1, LTF_(left),LTF_(right),−1,−1,x ₃ ,x ₄}where {x₁, x₂, x₃, x₄} can be determined based on PAPR. The {x₁, x₂, x₃,x₄} may be the best four values (1 or −1) to minimize the maximum PAPRsfor all possible pilot values. For example, {x₁, x₂, x₃, x₄} may beequal to {1, 1, −1, 1}.

The HE-LTF sequence for the 40 MHz and the 1×HE-LTF mode may includezero values on subcarrier indices from −244 to −4 excluding every foursubcarrier indices from −244 to −4, non-zero values on every foursubcarrier indices from −244 to −4, a zero value on a subcarrier indexof −3, DC tones on subcarrier indices from −2 to 2, a zero value on asubcarrier index of 3, non-zero values on every four subcarrier indicesfrom 4 to 244, and zero values on subcarrier indices from 4 to 244excluding every four subcarrier indices from 4 to 122.

For the HE-LTF sequence for the 40 MHz and the 1×HE-LTF mode,non-limiting examples of the sequences of non-zero values are asfollows.

In a first example, {y₁, y₂, y₃, y₄, LTF_(left), 1, LTF_(right), −1, −1,−1, 1, −1, 1, 1, −1, LTF_(left), 1, LTF_(right), y₅, y₆, y₇, y₈}; and

In a second example, {z₁, z₂, z₃, LTF_(left), 1, LTF_(right), −1, −1,−1, 1, z₄, z₅, −1, 1, 1, −1, LTF_(left), 1, LTF_(right), z₆, z₇, z₈}.

In the first and second examples, {y₁, y₂, y₃, y₄, y₅, y₆, y₇, y₈} and{z₁, z₂, z₃, z₄, z₅, z₆, z₇, z₈} can be determined based on PAPR. Theymay be the best values (1 or −1) to minimize the maximum PAPRs for allpossible pilot values. For example, {y₁, y₂, y₃, y₄, y₅, y₆, y₇, y₈} maybe equal to {1, −1, −1, 1, −1, −1, 1, 1}, and {z₁, z₂, z₃, z₄, z₅, z₆,z₇, z₈} may be equal to {1, 1, 1, −1, −1, 1, −1, −1}.

The HE-LTF sequence for the 80 MHz and the 1×HE-LTF mode may includezero values on subcarrier indices from −500 to −4 excluding every fourthsubcarrier index from −500 to −4, non-zero values on every fourthsubcarrier indices from −500 to −4, a zero value on a subcarrier indexof −3, DC tones on subcarrier indices from −2 to 2, a zero value on asubcarrier index of 3, non-zero values on every fourth subcarrier indexfrom 4 to 500, and zero values on subcarrier indices from 4 to 500excluding every fourth subcarrier index from 4 to 500.

For the HE-LTF sequence for the 80 MHz and the 1×HE-LTF mode,non-limiting examples of the sequences of non-zero values are asfollows.

In a first example, {p₁, p₂, p₃, p₄, LTF_(left), 1, LTF_(right), −1, −1,−1, 1, 1, −1, 1, −1, 1, 1, −1, LTF_(left), 1, LTF_(right), 1, −1, 1, −1,1, −1, −1, 1, LTF_(left), 1, LTF_(right), −1, −1, −1, 1, 1, −1, 1, −1,1, 1, −1, LTF_(left), 1, LTF_(right), p₅, p₆, p₇, p₈}; and

In a second example, {q₁, q₂, q₃, LTF_(left), 1, LTF_(right), −1, −1,−1, 1, 1, −1, 1, −1, 1, 1, −1, LTF_(left), 1, LTF_(right), 1, −1, 1, −1,q₄, q₅, 1, −1, −1, 1, LTF_(left), 1, LTF_(right), −1, −1, −1, 1, 1, −1,1, −1, 1, 1, −1, LTF_(left), 1, LTF_(right), q₆, q₇, q₈}.

In the first and second examples, {p₁, p₂, p₃, p₄, p₅, p₆, p₇, p₈} and{q₁, q₂, q₃, q₄, q₆, q₇, q₈} can be determined based on PAPR. They maybe the best values (1 or −1) to minimize the maximum PAPRs for allpossible pilot values. For example, {p₁, p₂, p₃, p₄, p₅, p₆, p₇, p₈} maybe equal to {−1, −1, −1, 1, −1, 1, −1, 1}, and {q₁, q₂, q₃, q₄, q₅, q₆,q₇, q₈} may be equal to {−1, 1, −1, 1, −1, 1, 1, −1}.

FIGS. 46 through 49 illustrate flow charts of examples of operationsfacilitating wireless communication.

FIG. 46 shows operations for facilitating transmission of a downlink HEPPDU by an AP station.

The AP station determines a channel bandwidth (S4601). The AP stationmay determine the channel bandwidth from among a plurality ofbandwidths. The plurality of bandwidths may include 20 MHz, 40 MHz, 80MHz, and 160 MHz.

The AP station determines an HE-LTF mode (S4603). The AP station maydetermine the HE-LTF mode from among a plurality of HE-LTF modes. Theplurality of HE-LTF modes may include a 4×HE-LTF mode, a 2×HE-LTF mode,and a 1×HE-LTF mode.

The AP station generates one or more LTF symbols (S4605). The AP stationmay generate the LTF symbols, by using an HE-LTF sequence correspondingto the determined channel bandwidth and the determined HE-LTF mode. Aduration of the LTF symbol excluding a guard interval can be 12.8microseconds when the determined HE-LTF mode is 4×HE-LTF mode. Aduration of the LTF symbol excluding a guard interval can be 6.4microseconds when the determined HE-LTF mode is 2×HE-LTF mode. Aduration of the LTF symbol excluding a guard interval can be 3.2microseconds when the determined HE-LTF mode is 1×HE-LTF mode.

The HE-LTF sequence used for generating the LTF symbol may be one of aplurality of HE-LTF sequences for the plurality of bandwidths and theplurality of HE-LTF modes. For example, the plurality of HE-LTFsequences may include a first HE-LTF sequence for 20 MHz channelbandwidth and the 4×HE-LTF mode, a second HE-LTF sequence for 20 MHzchannel bandwidth and the 2×HE-LTF mode, a third HE-LTF sequence for 40MHz channel bandwidth and the 4×HE-LTF mode, a fourth HE-LTF sequencefor 40 MHz channel bandwidth and the 2× HE-LTF mode, a fifth HE-LTFsequence for 80 MHz channel bandwidth and the 4×HE-LTF mode, a sixthHE-LTF sequence for 80 MHz channel bandwidth and the 2×HE-LTF mode, aseventh HE-LTF sequence for 20 MHz channel bandwidth and the 1×HE-LTFmode, an eighth HE-LTF sequence for 40 MHz channel bandwidth and the1×HE-LTF mode, a ninth HE-LTF sequence for 80 MHz channel bandwidth andthe 1×HE-LTF mode, a tenth HE-LTF sequence for 160 MHz channel bandwidthand the 4×HE-LTF mode, an eleventh HE-LTF sequence for 160 MHz channelbandwidth and the 2×HE-LTF mode, and a twelfth HE-LTF sequence for 160MHz channel bandwidth and the 1×HE-LTF mode. Various examples for thefirst through the twelfth HE-LTF sequences have been introduced above inthe present disclosure.

For example, the first HE-LTF sequence for non-OFDMA transmission mayinclude non-zero values on subcarrier indices from −122 to −2, DC toneson subcarrier indices from −1 to 1, and non-zero values on subcarrierindices from 2 to 122. For OFDMA transmission, the first HE-LTF sequencemay include zero values on subcarrier indices of −4, −3, 3, and 4.

The second HE-LTF sequence for non-OFDMA transmission may include zerovalues on every odd subcarrier index from −122 to −2, non-zero values onevery even subcarrier index from −122 to −2, DC tones on subcarrierindices from −1 to 1, non-zero values on every even subcarrier indexfrom 2 to 122, and zero values on every odd subcarrier index from 2 to122. For OFDMA transmission, the second HE-LTF sequence may include zerovalues on subcarrier indices of −3 and 3.

The third HE-LTF sequence for both non-OFDMA transmission and OFDMAtransmission may include non-zero values on subcarrier indices from −244to −3, DC tones on subcarrier indices from −2 to 2, and non-zero valueson subcarrier indices from 3 to 244. The third HE-LTF sequence mayinclude plural unit sequences for plural resource units, wherein each ofthe plural unit sequences corresponds to one of plural base sequencesmultiplied by one of +1 or −1. The number of the plural unit sequencescan be 18, the number of the plural resource units can be 18, and eachof the plural unit sequences may include 26 elements whose value are +1or −1. The number of the plural base sequences can be 2 or 4.

The fourth HE-LTF sequence for both non-OFDMA transmission and OFDMAtransmission may include zero values on every odd subcarrier index from−244 to −4, non-zero values on every even subcarrier index from −244 to−4, a zero value on a subcarrier index of −3, DC tones on subcarrierindices from −2 to 2, a zero value on a subcarrier index of 3, non-zerovalues on every even subcarrier index from 4 to 244, and zero values onevery odd subcarrier index from 4 to 244. The fourth HE-LTF sequence mayinclude plural unit sequences for plural resource units, wherein each ofthe plural unit sequences corresponds to one of plural base sequencesmultiplied by one of +1 or −1. The number of the plural unit sequencescan be 18, the number of the plural resource units can be 18, and eachof the plural unit sequences may include 26 elements whose value are +1or −1. The number of the plural base sequences can be 2 or 4.

The fifth HE-LTF sequence for non-OFDMA transmission may includenon-zero values on subcarrier indices from −500 to −3, DC tones onsubcarrier indices from −2 to 2, and non-zero values on subcarrierindices from 3 to 500. For OFDMA transmission, the fifth HE-LTF sequencemay include zero values on subcarrier indices of −3 and 3.

The sixth HE-LTF sequence for both non-OFDMA transmission and OFDMAtransmission may include zero values on every odd subcarrier index from−500 to −4, non-zero values on every even subcarrier index from −500 to−4, a zero value on a subcarrier index of −3, DC tones on subcarrierindices from −2 to 2, a zero value on a subcarrier index of 3, non-zerovalues on every even subcarrier index from 4 to 500, and zero values onevery odd subcarrier index from 4 to 500.

The seventh HE-LTF sequence for both non-OFDMA transmission and OFDMAtransmission may include zero values on subcarrier indices from −122 to−4 excluding every fourth subcarrier index from −122 to −4, non-zerovalues on every fourth subcarrier index from −122 to −4, zero values onsubcarrier indices from −3 to −2, DC tones on subcarrier indices from −1to 1, zero values on subcarrier indices of 2 and 3, non-zero values onevery fourth subcarrier index from 4 to 122, and zero values onsubcarrier indices from 4 to 122 excluding every fourth subcarrier indexfrom 4 to 122.

The eighth HE-LTF sequence for both non-OFDMA transmission and OFDMAtransmission may include zero values on subcarrier indices from −244 to−4 excluding every fourth subcarrier index from −244 to −4, non-zerovalues on every fourth subcarrier index from −244 to −4, a zero value onsubcarrier index of −3, DC tones on subcarrier indices from −2 to 2, azero value on subcarrier index of 3, non-zero values on every fourthsubcarrier index from 4 to 244, and zero values on subcarrier indicesfrom 4 to 244 excluding every fourth subcarrier index from 4 to 122.

The ninth HE-LTF sequence for both non-OFDMA transmission and OFDMAtransmission may include zero values on subcarrier indices from −500 to−4 excluding every fourth subcarrier index from −500 to −4, non-zerovalues on every fourth subcarrier index from −500 to −4, a zero value ona subcarrier index of −3, DC tones on subcarrier indices from −2 to 2, azero value on a subcarrier index of 3, non-zero values on every fourthsubcarrier index from 4 to 500, and zero values on subcarrier indicesfrom 4 to 500 excluding every fourth subcarrier index from 4 to 500.

The AP station transmits a downlink HE PPDU in the determined channelbandwidth (S4607). The HE-LTF field of the HE PPDU may include the oneor more LTF symbols which are generated in S4605. The format of thedownlink HE PPDU can follow that shown in FIG. 8 .

FIG. 47 shows operations for facilitating reception of a downlink HEPPDU by a non-AP station.

The non-AP station receives a downlink HE PPDU in a channel bandwidth(S4701). In an aspect, the downlink HE PPDU may be the downlink HE PPDUtransmitted by the AP station in S4607 of FIG. 46 . The HE-LTF field ofthe HE PPDU may include the one or more LTF symbols. The HE PPDU maycorrespond to a downlink HE PPDU. The channel bandwidth can be one of aplurality of bandwidths including 20 MHz, 40 MHz, 80 MHz, and 160 MHz.The format of the downlink HE PPDU can follow that shown in FIG. 8 .

The non-AP station obtains an HE-LTF sequence from the HE-LTF symbol(S4703). The HE-LTF sequence may correspond to the channel bandwidth andan HE-LTF mode of the HE-LTF symbol. The HE-LTF mode of the HE-LTFsymbol can be one of a plurality of HE-LTF modes including a 4×HE-LTFmode, a 2×HE-LTF mode, and a 1×HE-LTF mode. If the HE PPDU is a downlinknon-OFDMA HE PPDU or a downlink OFDMA HE PPDU, the non-AP station mayobtain the HE-LTF sequence on the whole subcarriers from the HE-LTFsymbol. If the HE PPDU is the downlink OFDMA HE PPDU, the non-AP stationmay obtain the HE-LTF sequence on a part of the whole subcarriers fromthe HE-LTF symbol. The part may correspond to a resource unit which isallocated to the non-AP station. In particular, subcarriers belonging tothe part may be the same as subcarriers belonging to the allocatedresource unit. Alternatively, subcarriers belonging to the part maycover subcarriers belonging to the allocated resource unit.

The non-AP station estimates a channel by using the HE-LTF sequence(S4705). If the HE PPDU is a downlink non-OFDMA HE PPDU or a downlinkOFDMA HE PPDU, the non-AP station may perform channel estimation on thewhole subcarriers. If the HE PPDU is the downlink OFDMA HE PPDU, thenon-AP station may perform channel estimation on the part of the wholesubcarriers.

FIG. 48 shows operations for facilitating transmission of an uplink HEPPDU by a non-AP station.

The non-AP station determines a channel bandwidth (S4801). The non-APstation may determine the channel bandwidth among a plurality ofbandwidths. The plurality of bandwidths may include 20 MHz, 40 MHz, 80MHz, and 160 MHz.

The non-AP station determines a resource unit that is allocated to thenon-AP station (S4802). The non-AP station may receive resourceallocation information for an uplink transmission, and determine theallocated resource unit based on the resource allocation information.

The non-AP station determines an HE-LTF mode (S4803). The non-AP stationmay determine the HE-LTF mode among a plurality of HE-LTF modes. Theplurality of HE-LTF modes may include a 4×HE-LTF mode, a 2×HE-LTF mode,and a 1×HE-LTF mode.

The non-AP station generates one or more LTF symbols (S4805). The non-APstation may generate the LTF symbols by using an HE-LTF sequencecorresponding to the determined channel bandwidth and the determinedHE-LTF mode. The non-AP station may use a part or the whole of theHE-LTF sequence. If the HE PPDU is an uplink non-OFDMA HE PPDU, thenon-AP station may use the whole of the HE-LTF sequence. If the HE PPDUis an uplink OFDMA HE PPDU, the non-AP station may use a part of theHE-LTF sequence. The part may correspond to a resource unit which isallocated to the non-AP station. In particular, subcarriers belonging tothe part may be the same as subcarriers belonging to the allocatedresource unit.

The non-AP station transmits an uplink HE PPDU to the AP station in thedetermined channel bandwidth (S4807). The HE-LTF field of the HE PPDUmay include the one or more LTF symbols which are generated in S4805.The format of the uplink HE PPDU can follow that shown in FIG. 8 .

FIG. 49 shows operations for facilitating reception of an uplink HE PPDUby an AP station.

The AP station receives an uplink HE PPDU in a channel bandwidth(S4901). In an aspect, the uplink HE PPDU may be the uplink HE PPDUtransmitted by the non-AP station in S4807 of FIG. 48 . The HE-LTF fieldof the HE PPDU may include the one or more LTF symbols. The HE PPDU maycorrespond to an uplink HE PPDU. The channel bandwidth can be one of aplurality of bandwidths including 20 MHz, 40 MHz, 80 MHz, and 160 MHz.The format of the uplink HE PPDU can follow that shown in FIG. 8 .

The AP station obtains an HE-LTF sequence from the HE-LTF symbol(S4903). The HE-LTF sequence may correspond to the channel bandwidth andan HE-LTF mode of the HE-LTF symbol.

The AP station estimates a channel by using the HE-LTF sequence (S4905).

FIGS. 50 through 52 illustrate flow charts of examples of operationsfacilitating wireless communication. For explanatory and illustrationpurposes, the example processes 5000, 5100 and 5200 may be performed bythe wireless communication devices 111-115 of FIG. 1 and theircomponents such as a baseband processor 210, a PHY processor 215, atransmitting signal processing unit 280 and/or a receiving signalprocessing unit 290; however, the example processes 5000, 5100 and 5200are not limited to the wireless communication devices 111-115 of FIG. 1or their components, and the example processes 5000, 5100 and 5200 maybe performed by other devices or components. Further for explanatory andillustration purposes, the blocks of the example processes 5000, 5100and 5200 are described herein as occurring in serial or linearly.However, multiple blocks of the example processes 5000, 5100 and 5200may occur in parallel. In addition, the blocks of the example processes5000, 5100 and 5200 need not be performed in the order shown and/or oneor more of the blocks or operations of the example processes 5000, 5100and 5200 need not be performed. One or more operations described withreference to FIGS. 50 through 52 may be performed as part of one or moreother operations associated thereof.

Various examples of aspects of the disclosure are described below asclauses for convenience. These are provided as examples, and do notlimit the subject technology. Any identifications of the figures andreference numbers are merely examples and for illustrative purposes, andthe clauses are not limited by those identifications. As an example,some of the clauses described below are illustrated in FIGS. 45 through52 .

Clause A: An apparatus for facilitating wireless communication, theapparatus comprising: one or more memories; and one or more processorscoupled to the one or more memories, wherein the one or more processorsare configured to cause: determining a channel bandwidth including a 20MHz channel bandwidth, 40 MHz channel bandwidth, and 80 MHz channelbandwidth; determining an HE-LTF mode among a plurality of HE-LTF modesincluding a 4×HE-LTF mode and a 2×HE-LTF mode; generating an HE-LTFsymbol by using an HE-LTF sequence corresponding to the determinedchannel bandwidth and the determined HE-LTF mode, wherein the HE-LTFsequence is among a plurality of HE-LTF sequences for the plurality ofbandwidths and the plurality of HE-LTF modes; and transmitting an HEPPDU including the HE-LTF symbol, in the determined channel bandwidth.

Clause B: An apparatus for facilitating wireless communication, theapparatus comprising: one or more memories; and one or more processorscoupled to the one or more memories, wherein the one or more processorsare configured to cause: receiving a downlink HE PPDU including anHE-LTF symbol, in a channel bandwidth which is one of a plurality ofbandwidths including 20 MHz, 40 MHz, and 80 MHz; and obtaining, from theHE-LTF symbol, an HE-LTF sequence corresponding to the channel bandwidthand an HE-LTF mode of the HE-LTF symbol, wherein the HE-LTF mode is oneof a plurality of HE-LTF modes including a 2×HE-LTF mode and a 4×HE-LTFmode.

In one or more aspects, the claims are incorporated herein as clauses.In one or more aspects, various phrases and sentences of this presentdisclosure are incorporated herein as clauses. Each of implementationsof Clauses A and B may include one or more of the clauses in anycombination or may include some portions of one or more of the clausesin any combination. In one or more aspects, some of the phrases or wordsin Clauses A and B may be changed or removed.

In one aspect, a method may be an operation, an instruction, or afunction and vice versa. In one aspect, a clause may depend from one ormore of other clauses. In one aspect, a clause may be amended to includesome or all of the words (e.g., instructions, operations, functions, orcomponents) in one or more other clauses, one or more sentences, one ormore phrases, one or more paragraphs, and/or one or more claims.

In one or more aspects, additional clauses are described below.

A method comprising one or more methods or operations of any one or moreof the clauses.

An apparatus comprising one or more memories and one or more processors(e.g., 210), the one or more processors configured to cause performingone or more methods or operations of any one or more of the clauses.

An apparatus comprising one or more memories (e.g., 240, one or moreinternal, external or remote memories, or one or more registers) and oneor more processors (e.g., 210) coupled to the one or more memories, theone or more processors configured to cause the apparatus to perform oneor more methods or operations of any one or more of the clauses.

A processor (e.g., 210) comprising modules for carrying out one or moremethods or operations of any one or more of the clauses.

A hardware apparatus comprising circuits (e.g., 210) configured toperform one or more methods or operations of any one or more of theclauses.

An apparatus comprising means (e.g., 210) adapted for performing one ormore methods or operations of any one or more of the clauses.

An apparatus comprising components (e.g., 210) operable to carry out oneor more methods or operations of any one or more of the clauses.

A computer-readable storage medium (e.g., 240, one or more internal,external or remote memories, or one or more registers) comprisinginstructions stored therein, the instructions comprising code forperforming one or more methods or operations of any one or more of theclauses.

A computer-readable storage medium (e.g., 240, one or more internal,external or remote memories, or one or more registers) storinginstructions that, when executed by one or more processors, cause one ormore processors to perform one or more methods or operations of any oneor more of the clauses.

To illustrate the interchangeability of hardware and software, itemssuch as the various illustrative blocks, modules, components, methods,operations, instructions, and algorithms have been described generallyin terms of their functionality. Whether such functionality isimplemented as hardware or software depends upon the particularapplication and design constraints imposed on the overall system.Skilled artisans may implement the described functionality in varyingways for each particular application.

A reference to an element in the singular is not intended to mean oneand only one unless specifically so stated, but rather one or more. Forexample, “a” module may refer to one or more modules. An elementproceeded by “a,” “an,” “the,” or “said” does not, without furtherconstraints, preclude the existence of additional same elements.

Headings and subheadings, if any, are used for convenience only and donot limit the invention. The word exemplary is used to mean serving asan example or illustration. To the extent that the term “include,”“have,” or the like is used, such term is intended to be inclusive in amanner similar to the term “comprise” as “comprise” is interpreted whenemployed as a transitional word in a claim. Relational terms such asfirst and second and the like may be used to distinguish one entity oraction from another without necessarily requiring or implying any actualsuch relationship or order between such entities or actions.

Phrases such as an aspect, the aspect, another aspect, some aspects, oneor more aspects, an implementation, the implementation, anotherimplementation, some implementations, one or more implementations, anembodiment, the embodiment, another embodiment, some embodiments, one ormore embodiments, a configuration, the configuration, anotherconfiguration, some configurations, one or more configurations, thesubject technology, the disclosure, the present disclosure, othervariations thereof and alike are for convenience and do not imply that adisclosure relating to such phrase(s) is essential to the subjecttechnology or that such disclosure applies to all configurations of thesubject technology. A disclosure relating to such phrase(s) may apply toall configurations, or one or more configurations. A disclosure relatingto such phrase(s) may provide one or more examples. A phrase such as anaspect or some aspects may refer to one or more aspects and vice versa,and this applies similarly to other foregoing phrases.

A phrase “at least one of” preceding a series of items, with the terms“and” or “or” to separate any of the items, modifies the list as awhole, rather than each member of the list. The phrase “at least one of”does not require selection of at least one item; rather, the phraseallows a meaning that includes at least one of any one of the items,and/or at least one of any combination of the items, and/or at least oneof each of the items. By way of example, each of the phrases “at leastone of A, B, and C” or “at least one of A, B, or C” refers to only A,only B, or only C; any combination of A, B, and C; and/or at least oneof each of A, B, and C.

It is understood that the specific order or hierarchy of steps,operations, or processes disclosed is an illustration of exemplaryapproaches. Unless explicitly stated otherwise, it is understood thatthe specific order or hierarchy of steps, operations, or processes maybe performed in different order. Some of the steps, operations, orprocesses may be performed simultaneously or may be performed as a partof one or more other steps, operations, or processes. The accompanyingmethod claims, if any, present elements of the various steps, operationsor processes in a sample order, and are not meant to be limited to thespecific order or hierarchy presented. These may be performed in serial,linearly, in parallel or in different order. It should be understoodthat the described instructions, operations, and systems can generallybe integrated together in a single software/hardware product or packagedinto multiple software/hardware products.

The disclosure is provided to enable any person skilled in the art topractice the various aspects described herein. In some instances,well-known structures and components are shown in block diagram form inorder to avoid obscuring the concepts of the subject technology. Thedisclosure provides various examples of the subject technology, and thesubject technology is not limited to these examples. Variousmodifications to these aspects will be readily apparent to those skilledin the art, and the principles described herein may be applied to otheraspects.

All structural and functional equivalents to the elements of the variousaspects described throughout this disclosure that are known or latercome to be known to those of ordinary skill in the art are expresslyincorporated herein by reference and are intended to be encompassed bythe claims. Moreover, nothing disclosed herein is intended to bededicated to the public regardless of whether such disclosure isexplicitly recited in the claims. No claim element is to be construedunder the provisions of 35 U.S.C. § 112, sixth paragraph, unless theelement is expressly recited using a phrase means for or, in the case ofa method claim, the element is recited using the phrase step for.

The title, background, brief description of the drawings, abstract, anddrawings are hereby incorporated into the disclosure and are provided asillustrative examples of the disclosure, not as restrictivedescriptions. It is submitted with the understanding that they will notbe used to limit the scope or meaning of the claims. In addition, in thedetailed description, it can be seen that the description providesillustrative examples and the various features are grouped together invarious implementations for the purpose of streamlining the disclosure.This method of disclosure is not to be interpreted as reflecting anintention that the claimed subject matter requires more features thanare expressly recited in each claim. Rather, as the following claimsreflect, inventive subject matter lies in less than all features of asingle disclosed configuration or operation. The following claims arehereby incorporated into the detailed description, with each claimstanding on its own as a separately claimed subject matter.

The claims are not intended to be limited to the aspects describedherein, but are to be accorded the full scope consistent with thelanguage claims and to encompass all legal equivalents. Notwithstanding,none of the claims are intended to embrace subject matter that fails tosatisfy the requirements of the applicable patent law, nor should theybe interpreted in such a way.

What is claimed is:
 1. An apparatus for facilitating wirelesscommunication, the apparatus comprising: one or more memories comprisinginstructions; and one or more processors coupled to the one or morememories, wherein when the one or more processors execute theinstructions, the apparatus performs operations comprising: selecting,based on a bandwidth of a frame, a high efficiency long training field(HE-LTF) sequence for the frame; transmitting, to a set of stations,only a set of allocated subcarriers of the HE-LTF sequence for theframe, wherein each subcarrier in the set of allocated subcarriers isallocated to a station in the set of stations for the frame; andtransmitting, to the set of stations, a set of additional fields of theframe using the set of allocated subcarriers and a set of unallocatedsubcarriers, the set of additional fields including a legacy shorttraining field (L-STF), a legacy long training field (L-LTF), a legacysignaling field (L-SIG), a high efficiency signaling A field (HE-SIG-A),and a high efficiency short training field (HE-STF), wherein the set ofunallocated subcarriers in the bandwidth of the frame is not allocatedto a station for the frame and no energy is transmitted on the set ofunallocated subcarriers during transmission of the HE-LTF sequence forthe frame.
 2. The apparatus of claim 1, wherein the set of unallocatedsubcarriers includes data subcarriers.
 3. The apparatus of claim 1,wherein when the one or more processors execute the instructions, theapparatus performs the operations comprising: transmitting the framesuch that the set of unallocated subcarriers in the HE-LTF sequence havea value of zero.
 4. The apparatus of claim 1, wherein the frame is partof a downlink orthogonal frequency division multiple access (OFDMA)transmission and the apparatus is an access point, and wherein when theone or more processors execute the instructions, the access pointperforms the operations comprising: determining allocation ofsubcarriers to the set of stations.
 5. The apparatus of claim 1, whereinthe frame is part of an uplink orthogonal frequency division multipleaccess (OFDMA) transmission and the apparatus is a non-access pointstation, and wherein the set of stations includes an access point thatsets the bandwidth of the frame.
 6. The apparatus of claim 1, whereinthe bandwidth of the frame is one of 20 MHz, 40 MHz, 80 MHz, 160 MHz, or80+80 MHz.
 7. A method for facilitating wireless communication, themethod comprising: selecting, by a wireless device based on a bandwidthof a frame, a high efficiency long training field (HE-LTF) sequence forthe frame; transmitting, by the wireless device to a set of stations,only a set of allocated subcarriers of the HE-LTF sequence for theframe, wherein each subcarrier in the set of allocated subcarriers isallocated to a station in the set of stations for the frame; andtransmitting, by the wireless device to the set of stations, a set ofadditional fields of the frame using the set of allocated subcarriersand a set of unallocated subcarriers, the set of additional fieldsincluding a legacy short training field (L-STF), a legacy long trainingfield (L-LTF), a legacy signaling field (L-SIG), a high efficiencysignaling A field (HE-SIG-A), and a high efficiency short training field(HE-STF), wherein the set of unallocated subcarriers in the bandwidth ofthe frame is not allocated to a station for the frame and no energy istransmitted on the set of unallocated subcarriers during transmission ofthe HE-LTF sequence for the frame.
 8. The method of claim 7, wherein theset of unallocated subcarriers includes data subcarriers.
 9. The methodof claim 7, comprising: transmitting the frame such that the set ofunallocated subcarriers in the HE-LTF sequence have a value of zero. 10.The method of claim 7, wherein the frame is part of a downlinkorthogonal frequency division multiple access (OFDMA) transmission, andwherein the method comprises: determining allocation of subcarriers tothe set of stations.
 11. The method of claim 7, wherein the frame ispart of an uplink orthogonal frequency division multiple access (OFDMA)transmission and the apparatus is a non-access point station, andwherein the set of stations includes an access point that sets thebandwidth of the frame.
 12. The method of claim 7, wherein the bandwidthof the frame is one of 20 MHz, 40 MHz, 80 MHz, 160 MHz, or 80+80 MHz.13. A non-transitory machine-readable storage medium that storesinstructions, which when executed by one or more processors of awireless device, cause the wireless device to: select, based on abandwidth of a frame, a high efficiency long training field (HE-LTF)sequence for the frame; transmit, to a set of stations, only a set ofallocated subcarriers of the HE-LTF sequence for the frame, wherein eachsubcarrier in the set of allocated subcarriers is allocated to a stationin the set of stations for the frame; and transmit, to the set ofstations, a set of additional fields of the frame using the set ofallocated subcarriers and a set of unallocated subcarriers, the set ofadditional fields includes a legacy short training field (L-STF), alegacy long training field (L-LTF), a legacy signaling field (L-SIG), ahigh efficiency signaling A field (HE-SIG-A), and a high efficiencyshort training field (HE-STF), wherein the set of unallocatedsubcarriers in the bandwidth of the frame is not allocated to a stationfor the frame and no energy is transmitted on the set of unallocatedsubcarriers during transmission of the HE-LTF sequence for the frame.14. The non-transitory machine-readable storage medium of claim 13,wherein the set of unallocated subcarriers includes data subcarriers.15. The non-transitory machine-readable storage medium of claim 13,wherein when the one or more processors execute the instructions, thewireless device is caused to perform operations comprising: transmittingthe frame such that the set of unallocated subcarriers in the HE-LTFsequence have a value of zero.
 16. The non-transitory machine-readablestorage medium of claim 13, wherein the frame is part of a downlinkorthogonal frequency division multiple access (OFDMA) transmission andthe wireless device is an access point, and wherein when the one or moreprocessors execute the instructions, the access point performsoperations comprising: determining allocation of subcarriers to the setof stations.
 17. The non-transitory machine-readable storage medium ofclaim 13, wherein the frame is part of an uplink orthogonal frequencydivision multiple access (OFDMA) transmission and the apparatus is anon-access point station, and wherein the set of stations includes anaccess point that sets the bandwidth of the frame.